使用php将数据插入数据库时出现问题
我在将数据插入数据库时遇到了问题，尽管它已成功插入，但插入数据库的数据会变成整数使用php将数据插入数据库时出现问题,php,mysql,Php,Mysql,我在将数据插入数据库时遇到了问题，尽管它已成功插入，但插入数据库的数据会变成整数 <?php //connect to database include "databaseconnection.php"; ?> <body bgcolor="#e5edf8"> <form name="addform" method="post" action="../insert.php"> <table width="1110" height="184"
<?php
//connect to database
include "databaseconnection.php";
?>
<body bgcolor="#e5edf8">
<form name="addform" method="post" action="../insert.php">
<table width="1110" height="184" border="0" cellpadding="0" cellspacing="0">
<tr>
<td width="393" height="28"><strong> TICKET NO :
<?php
$sql = mysql_query("SELECT * FROM troubleticket");
$record = mysql_fetch_array($sql);
echo '<input type="text" name="ticketnofld" readonly value=" '.$record['ticketno'].' " >';
?>
</strong>
</td>
<td width="717"> <strong>TECHNICAL NAME :
<?php
echo "<select name='techname' type='text'>";
echo '<option id="0">'.'--Select technical Name--'.'</option>';
$sql = mysql_query("SELECT * FROM technical");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['FNAME']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td><strong>COMPANY NAME :
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
</select>
</strong></td>
<td><strong>TYPE OF SERVICE :
<?php
echo " <select name='typeofservice' type='text'>";
echo '<option id="0">'.'--Select type of service--'.'</option>';
$sql = mysql_query("SELECT * FROM typeofservice");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['tosid'].'">'.$record['typeofservice']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td colspan="2"><p><strong>PROBLEM :<textarea type='text' rows="10" cols="100" name="problemfld" id="problemfld"> </textarea>
</strong></p>
</td>
</tr>
<tr>
<td colspan="2"><input type="submit" name="Addbutton" value="ADD" >
<input type="button" name="Cancelbutton" value="CANCEL" ></td>
</tr>
</table>
</form>
票号：
技术名称：
我在将数据插入数据库时遇到问题，尽管它是
已成功插入，但随后将数据插入数据库
变成整数
<?php
//connect to database
include "databaseconnection.php";
?>
<body bgcolor="#e5edf8">
<form name="addform" method="post" action="../insert.php">
<table width="1110" height="184" border="0" cellpadding="0" cellspacing="0">
<tr>
<td width="393" height="28"><strong> TICKET NO :
<?php
$sql = mysql_query("SELECT * FROM troubleticket");
$record = mysql_fetch_array($sql);
echo '<input type="text" name="ticketnofld" readonly value=" '.$record['ticketno'].' " >';
?>
</strong>
</td>
<td width="717"> <strong>TECHNICAL NAME :
<?php
echo "<select name='techname' type='text'>";
echo '<option id="0">'.'--Select technical Name--'.'</option>';
$sql = mysql_query("SELECT * FROM technical");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['FNAME']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td><strong>COMPANY NAME :
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
</select>
</strong></td>
<td><strong>TYPE OF SERVICE :
<?php
echo " <select name='typeofservice' type='text'>";
echo '<option id="0">'.'--Select type of service--'.'</option>';
$sql = mysql_query("SELECT * FROM typeofservice");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['tosid'].'">'.$record['typeofservice']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td colspan="2"><p><strong>PROBLEM :<textarea type='text' rows="10" cols="100" name="problemfld" id="problemfld"> </textarea>
</strong></p>
</td>
</tr>
<tr>
<td colspan="2"><input type="submit" name="Addbutton" value="ADD" >
<input type="button" name="Cancelbutton" value="CANCEL" ></td>
</tr>
</table>
</form>
是的，因为这就是您的
标记中的内容。例如，“公司名称”的选择列表如下：
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
注意我是如何将
更改为
并设置'
此外，您还有：
$techname=isset($_POST['techname']);
$companyname=isset($_POST['companyname']);
$typeofservice=isset($_POST['typeofservice']);
$problem=isset($_POST['problemfld']);
而isset
只返回0
或1
。因此，它只是检查$\u POST
值是否存在，而不是检查其他值。要快速测试，只需将其更改为：
$techname=$_POST['techname'];
$companyname=$_POST['companyname'];
$typeofservice=$_POST['typeofservice'];
$problem=$_POST['problemfld'];
但考虑到代码的重复性，我建议将其压缩为：
// Set the database connection.
$con = mysql_connect("127.0.0.1","root","","ojt") or die('Could not reach the database'.mysql_error());
// Set an array of post values.
$post_array = array('techname', 'companyname', 'typeofservice', 'problemfld');
// Roll through the post values, validate & assign them.
foreach ($post_array as $post_key => $post_value) {
$$post_key = '';
if (isset($_POST[$post_key])) {
$$post_key = mysql_real_escape_string(stripslashes($_POST[$post_key]));
}
}
// Set the query.
$sql = "INSERT INTO `ojt`.`troubleticket` (`ticketno`, `technicalname`, `services`, `problem`, `companyname`, `remarks`) VALUES (' ','$techname', '$typeofservice', '$problemfld', '$companyname', 'NEW')";
// Run the query.
$query = mysql_query($sql,$con);
// Check if the query ran.
if ($query) {
echo '1 Data Added';
}
else {
echo 'Unsuccessfully Saved';
}
是的，mysql.*
扩展在PHP5.3和5.4中被贬低，在5.5版中将被删除，因此您应该了解mysqli.*
的用法。这类似于mysql，但由您来处理
我在将数据插入数据库时遇到问题，尽管它是
已成功插入，但随后将数据插入数据库
变成整数
<?php
//connect to database
include "databaseconnection.php";
?>
<body bgcolor="#e5edf8">
<form name="addform" method="post" action="../insert.php">
<table width="1110" height="184" border="0" cellpadding="0" cellspacing="0">
<tr>
<td width="393" height="28"><strong> TICKET NO :
<?php
$sql = mysql_query("SELECT * FROM troubleticket");
$record = mysql_fetch_array($sql);
echo '<input type="text" name="ticketnofld" readonly value=" '.$record['ticketno'].' " >';
?>
</strong>
</td>
<td width="717"> <strong>TECHNICAL NAME :
<?php
echo "<select name='techname' type='text'>";
echo '<option id="0">'.'--Select technical Name--'.'</option>';
$sql = mysql_query("SELECT * FROM technical");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['FNAME']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td><strong>COMPANY NAME :
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
</select>
</strong></td>
<td><strong>TYPE OF SERVICE :
<?php
echo " <select name='typeofservice' type='text'>";
echo '<option id="0">'.'--Select type of service--'.'</option>';
$sql = mysql_query("SELECT * FROM typeofservice");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['tosid'].'">'.$record['typeofservice']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td colspan="2"><p><strong>PROBLEM :<textarea type='text' rows="10" cols="100" name="problemfld" id="problemfld"> </textarea>
</strong></p>
</td>
</tr>
<tr>
<td colspan="2"><input type="submit" name="Addbutton" value="ADD" >
<input type="button" name="Cancelbutton" value="CANCEL" ></td>
</tr>
</table>
</form>
是的，因为这就是您的
标记中的内容。例如，“公司名称”的选择列表如下：
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
注意我是如何将
更改为
并设置'
此外，您还有：
$techname=isset($_POST['techname']);
$companyname=isset($_POST['companyname']);
$typeofservice=isset($_POST['typeofservice']);
$problem=isset($_POST['problemfld']);
而isset
只返回0
或1
。因此，它只是检查$\u POST
值是否存在，而不是检查其他值。要快速测试，只需将其更改为：
$techname=$_POST['techname'];
$companyname=$_POST['companyname'];
$typeofservice=$_POST['typeofservice'];
$problem=$_POST['problemfld'];
但考虑到代码的重复性，我建议将其压缩为：
// Set the database connection.
$con = mysql_connect("127.0.0.1","root","","ojt") or die('Could not reach the database'.mysql_error());
// Set an array of post values.
$post_array = array('techname', 'companyname', 'typeofservice', 'problemfld');
// Roll through the post values, validate & assign them.
foreach ($post_array as $post_key => $post_value) {
$$post_key = '';
if (isset($_POST[$post_key])) {
$$post_key = mysql_real_escape_string(stripslashes($_POST[$post_key]));
}
}
// Set the query.
$sql = "INSERT INTO `ojt`.`troubleticket` (`ticketno`, `technicalname`, `services`, `problem`, `companyname`, `remarks`) VALUES (' ','$techname', '$typeofservice', '$problemfld', '$companyname', 'NEW')";
// Run the query.
$query = mysql_query($sql,$con);
// Check if the query ran.
if ($query) {
echo '1 Data Added';
}
else {
echo 'Unsuccessfully Saved';
}
是的，mysql.*
扩展在PHP5.3和5.4中被贬低，在5.5版中将被删除，因此您应该了解mysqli.*
的用法。这类似于mysql，但由您来处理
我在将数据插入数据库时遇到问题，尽管它是
已成功插入，但随后将数据插入数据库
变成整数
<?php
//connect to database
include "databaseconnection.php";
?>
<body bgcolor="#e5edf8">
<form name="addform" method="post" action="../insert.php">
<table width="1110" height="184" border="0" cellpadding="0" cellspacing="0">
<tr>
<td width="393" height="28"><strong> TICKET NO :
<?php
$sql = mysql_query("SELECT * FROM troubleticket");
$record = mysql_fetch_array($sql);
echo '<input type="text" name="ticketnofld" readonly value=" '.$record['ticketno'].' " >';
?>
</strong>
</td>
<td width="717"> <strong>TECHNICAL NAME :
<?php
echo "<select name='techname' type='text'>";
echo '<option id="0">'.'--Select technical Name--'.'</option>';
$sql = mysql_query("SELECT * FROM technical");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['FNAME']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td><strong>COMPANY NAME :
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
</select>
</strong></td>
<td><strong>TYPE OF SERVICE :
<?php
echo " <select name='typeofservice' type='text'>";
echo '<option id="0">'.'--Select type of service--'.'</option>';
$sql = mysql_query("SELECT * FROM typeofservice");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['tosid'].'">'.$record['typeofservice']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td colspan="2"><p><strong>PROBLEM :<textarea type='text' rows="10" cols="100" name="problemfld" id="problemfld"> </textarea>
</strong></p>
</td>
</tr>
<tr>
<td colspan="2"><input type="submit" name="Addbutton" value="ADD" >
<input type="button" name="Cancelbutton" value="CANCEL" ></td>
</tr>
</table>
</form>
是的，因为这就是您的
标记中的内容。例如，“公司名称”的选择列表如下：
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
注意我是如何将
更改为
并设置'
此外，您还有：
$techname=isset($_POST['techname']);
$companyname=isset($_POST['companyname']);
$typeofservice=isset($_POST['typeofservice']);
$problem=isset($_POST['problemfld']);
而isset
只返回0
或1
。因此，它只是检查$\u POST
值是否存在，而不是检查其他值。要快速测试，只需将其更改为：
$techname=$_POST['techname'];
$companyname=$_POST['companyname'];
$typeofservice=$_POST['typeofservice'];
$problem=$_POST['problemfld'];
但考虑到代码的重复性，我建议将其压缩为：
// Set the database connection.
$con = mysql_connect("127.0.0.1","root","","ojt") or die('Could not reach the database'.mysql_error());
// Set an array of post values.
$post_array = array('techname', 'companyname', 'typeofservice', 'problemfld');
// Roll through the post values, validate & assign them.
foreach ($post_array as $post_key => $post_value) {
$$post_key = '';
if (isset($_POST[$post_key])) {
$$post_key = mysql_real_escape_string(stripslashes($_POST[$post_key]));
}
}
// Set the query.
$sql = "INSERT INTO `ojt`.`troubleticket` (`ticketno`, `technicalname`, `services`, `problem`, `companyname`, `remarks`) VALUES (' ','$techname', '$typeofservice', '$problemfld', '$companyname', 'NEW')";
// Run the query.
$query = mysql_query($sql,$con);
// Check if the query ran.
if ($query) {
echo '1 Data Added';
}
else {
echo 'Unsuccessfully Saved';
}
是的，mysql.*
扩展在PHP5.3和5.4中被贬低，在5.5版中将被删除，因此您应该了解mysqli.*
的用法。这类似于mysql，但由您来处理
我在将数据插入数据库时遇到问题，尽管它是
已成功插入，但随后将数据插入数据库
变成整数
<?php
//connect to database
include "databaseconnection.php";
?>
<body bgcolor="#e5edf8">
<form name="addform" method="post" action="../insert.php">
<table width="1110" height="184" border="0" cellpadding="0" cellspacing="0">
<tr>
<td width="393" height="28"><strong> TICKET NO :
<?php
$sql = mysql_query("SELECT * FROM troubleticket");
$record = mysql_fetch_array($sql);
echo '<input type="text" name="ticketnofld" readonly value=" '.$record['ticketno'].' " >';
?>
</strong>
</td>
<td width="717"> <strong>TECHNICAL NAME :
<?php
echo "<select name='techname' type='text'>";
echo '<option id="0">'.'--Select technical Name--'.'</option>';
$sql = mysql_query("SELECT * FROM technical");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['FNAME']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td><strong>COMPANY NAME :
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
</select>
</strong></td>
<td><strong>TYPE OF SERVICE :
<?php
echo " <select name='typeofservice' type='text'>";
echo '<option id="0">'.'--Select type of service--'.'</option>';
$sql = mysql_query("SELECT * FROM typeofservice");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['tosid'].'">'.$record['typeofservice']. '</option>';
}
echo '</select>';
?>
</strong></td>
</tr>
<tr>
<td colspan="2"><p><strong>PROBLEM :<textarea type='text' rows="10" cols="100" name="problemfld" id="problemfld"> </textarea>
</strong></p>
</td>
</tr>
<tr>
<td colspan="2"><input type="submit" name="Addbutton" value="ADD" >
<input type="button" name="Cancelbutton" value="CANCEL" ></td>
</tr>
</table>
</form>
是的，因为这就是您的
标记中的内容。例如，“公司名称”的选择列表如下：
<?php
echo "<select name='companyname' type='text'>";
echo '<option id="0">'.'--Select Company Name--'.'</option>';
$sql = mysql_query("SELECT * FROM client");
while($record = mysql_fetch_assoc($sql))
{
echo '<option value=" '.$record['ID'].'">'.$record['NAME']. '</option>';
}
echo '</select>';
?>
注意我是如何将
更改为
并设置'
此外，您还有：
$techname=isset($_POST['techname']);
$companyname=isset($_POST['companyname']);
$typeofservice=isset($_POST['typeofservice']);
$problem=isset($_POST['problemfld']);
而isset
只返回0
或1
。因此，它只是检查$\u POST
值是否存在，而不是检查其他值。要快速测试，只需将其更改为：
$techname=$_POST['techname'];
$companyname=$_POST['companyname'];
$typeofservice=$_POST['typeofservice'];
$problem=$_POST['problemfld'];
但考虑到代码的重复性，我建议将其压缩为：
// Set the database connection.
$con = mysql_connect("127.0.0.1","root","","ojt") or die('Could not reach the database'.mysql_error());
// Set an array of post values.
$post_array = array('techname', 'companyname', 'typeofservice', 'problemfld');
// Roll through the post values, validate & assign them.
foreach ($post_array as $post_key => $post_value) {
$$post_key = '';
if (isset($_POST[$post_key])) {
$$post_key = mysql_real_escape_string(stripslashes($_POST[$post_key]));
}
}
// Set the query.
$sql = "INSERT INTO `ojt`.`troubleticket` (`ticketno`, `technicalname`, `services`, `problem`, `companyname`, `remarks`) VALUES (' ','$techname', '$typeofservice', '$problemfld', '$companyname', 'NEW')";
// Run the query.
$query = mysql_query($sql,$con);
// Check if the query ran.
if ($query) {
echo '1 Data Added';
}
else {
echo 'Unsuccessfully Saved';
}
是的，mysql.*
扩展在PHP5.3和5.4中被贬低，在5.5版中将被删除，因此您应该了解mysqli.*
的用法。这类似于mysql，但由您来处理。如果您存储的是整数而不是数据，请检查以下代码：
$techname=isset($_POST['techname']);
$companyname=isset($_POST['companyname']);
$typeofservice=isset($_POST['typeofservice']);
$problem=isset($_POST['problemfld']);
它将变量设置为1
，作为isset
您需要将其更改为：
$techname=isset($_POST['techname']) ? $_POST['techname'] : '';
$companyname=isset($_POST['companyname']) ? $_POST['companyname'] : '';
$typeofservice=isset($_POST['typeofservice']) ? $_POST['typeofservice'] : '';
$problem=isset($_POST['problemfld']) ? $_POST['problemfld'] : '';
如果存储的是整数而不是数据，请检查以下代码：
$techname=isset($_POST['techname']);
$companyname=isset($_POST['companyname']);
$typeofservice=isset($_POST['typeofservice']);
$problem=isset($_POST['problemfld']);
它将变量设置为1
，作为isset
您需要将其更改为：
$techname=isset($_POST['techname']) ? $_POST['techname'] : '';
$companyname=isset($_POST['companyname']) ? $_POST['companyname'] : '';
$typeofservice=isset($_POST['typeofservice']) ? $_POST['typeofservice'] : '';
$problem=isset($_POST['problemfld']) ? $_POST['problemfld'] : '';
如果存储的是整数而不是数据，请检查以下代码：
$techname=isset($_POST['techname']);
$companyname=isset($_POST['companyname']);
$typeofservice=isset($_POST['typeofservice']);
$problem=isset($_POST['problemfld']);
它将变量设置为1
，作为isset
您需要将其更改为：
$techname=isset($_POST['techname']) ? $_POST['techname'] : '';
$companyname=isset($_POST['companyname']) ? $_POST['companyname'] : '';
$typeofservice=isset($_POST['typeofservice']) ? $_POST['typeofservice'] : '';
$problem=isset($_POST['problemfld']) ? $_POST['problemfld'] : '';
如果存储的是整数而不是数据，请检查以下代码：
$techname=isset($_POST['techname']);
$companyname=isset($_POST['companyname']);
$typeofservice=isset($_POST['typeofservice']);
$problem=isset($_POST['problemfld']);
它将变量设置为1
，作为isset
您需要将其更改为：
$techname=isset($_POST['techname']) ? $_POST['techname'] : '';
$companyname=isset($_POST['companyname']) ? $_POST['companyname'] : '';
$typeofservice=isset($_POST['typeofservice']) ? $_POST['typeofservice'] : '';
$problem=isset($_POST['problemfld']) ? $_POST['problemfld'] : '';
您的数据库结构是什么？你确信字段有文本吗？MySQL在PHP 5.5。x中被贬低，考虑切换到myQuiLi。在<代码> > $TealMe= ISSET（$POST（'TealMe]））中肯定有与<代码> ISSET <代码>有关的内容；code>etc.@Fred ii-isset
返回一个布尔值，正确的方法是，if（isset（$\u POST['techname'））$techname=$\u POST['techname'）这整件事本来可以很容易避免，包括在使用PDO或准备语句时进行重写。您的数据库结构是什么？你确信字段有文本吗？MySQL在PHP 5.5。x中被贬低，考虑切换到myQuiLi。在<代码> > $TealMe= ISSET（$POST（'TealMe]））中肯定有与<代码> ISSET <代码>有关的内容；code>etc.@Fred ii-isset
返回一个布尔值，正确的方法是，if（isset（$\u POST['techname'））$techname=$\u POST['techname'）这整件事本来可以很容易避免，包括在使用PDO或准备语句时进行重写。您的数据库结构是什么？你确信字段有文本吗？MySQL在PHP 5.5。x中被贬低，考虑切换到myQuLi。肯定与