如何用PHP构建动态MySQL INSERT语句
你好如何用PHP构建动态MySQL INSERT语句,php,mysql,post,show,implode,Php,Mysql,Post,Show,Implode,你好 表单的这一部分显示mysql表中的列名称（计算机上安装的应用程序的名称），并使用YES/NO选项或input type=“text”框创建一个表单，以获得应用程序的其他权限 如何使用POST和mysql_query insert-INTO将其插入到mysql表中 列的数量正在更改，因为有另一个表单用于添加具有/不具有权限的应用程序 <tr bgcolor=#ddddff>'; //mysql_query for getting columns names $result =
表单的这一部分显示mysql表中的列名称（计算机上安装的应用程序的名称），并使用YES/NO选项或input type=“text”框创建一个表单，以获得应用程序的其他权限
如何使用POST和mysql_query insert-INTO将其插入到mysql表中
列的数量正在更改，因为有另一个表单用于添加具有/不具有权限的应用程序
<tr bgcolor=#ddddff>';
//mysql_query for getting columns names
$result = mysql_query("SHOW COLUMNS FROM employees") or die(mysql_error());
while ($row = mysql_fetch_array($result))
{
//exclude these columns bcs these are in other part of form
if($row[0] == 'id' || $row[0] == 'nameandsurname' || $row[0] == 'department'
|| $row[0] == 'phone' || $row[0] == 'computer' || $row[0] == 'data')
continue;
echo '<td bgcolor=#ddddff>'.$row[0].'<br />';
if (stripos($row[0], "privileges") !== false) {
echo '<td bgcolor=#ddddff><p><a class=hint href=#>
<input type="text" name="'.$row[0].'">
<span>Privileges like "occupation" or "like someone"</span></a></p></td></tr>';
}
else
{
echo '<td bgcolor=#ddddff align=center><select name="'.$row[0].'">
<option value = "No">No
<option value = "Yes">Yes
</td>
</tr>';
}
}
trim($_POST); // ????
$query = "INSERT INTO 'employees' VALUES (??)"; // ????
”；
//用于获取列名称的mysql\u查询
$result=mysql_query（“显示来自员工的列”）或die（mysql_error（））；
while（$row=mysql\u fetch\u数组（$result））
{
//排除这些列BC这些列在表单的其他部分
如果（$row[0]='id'| |$row[0]=='name和姓氏'| |$row[0]=='department'
||$row[0]=='phone'| |$row[0]=='computer'| |$row[0]=='data'）
继续；
回显“.$row[0]”。
；
if（stripos（$row[0]，“privileges”）！==false）{
回声“”；
}
其他的
{
回声'
不
对
';
}
}
修剪（$_POST）；/？？？？
$query=“插入“员工”值（？？）；/？？”？？？？
$exclude = array("id", "nameandsurname", "departument", "phone", "computer", "date");
$result = mysql_query("SHOW COLUMNS FROM employees") or die(mysql_error());
$columns = array();
while ($row = mysql_fetch_array($result)) {
if (!in_array($row[0], $exclude) {
$columns[] = $row[0];
}
}
$columnsforeach ($columns as $column) {
echo '<tr><td bgcolor="#ddddff">'.$column.'<br />';
if (stripos($column, "privileges") !== false) {
echo '<p><a class="hint" href="#">
<input type="text" name="'.$column.'">
<span>Privileges like "occupation" or "like someone"</span></a>';
} else {
echo '<select name="'.$column.'">
<option value = "No">No
<option value = "Yes">Yes
</select>';
}
echo '</td></tr>';
}
INFORMATION\u SCHEMA.COLUMNS
<html>
<body>
<form action="dynamicinsert.php" method="POST" >
user name:<br>
<input type="text" id="username" name="username">
<br><br>
first name:<br>
<input type="text" id="firstname" name="firstname">
<br><br>
password:<br>
<input type="password" id="password" name="password">
<br><br>
<input type="submit" name="submit" value="add" />
</form>
</body>
</html>
<?php
$servername = "localhost";
$username = "your_username";
$password = "your_password";
$dbname = "you_DB_name";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
function insertqueryfunction($dbfield,$table) {
$count = 0;
$fields = '';
foreach($dbfield as $col => $val) {
if ($count++ != 0) $fields .= ', ';
$col = addslashes($col);
$val = addslashes($val);
$fields .= "`$col` = '$val'";
}
$query = "INSERT INTO $table SET $fields;";
return $query;
}
if(isset($_POST['submit']))
{
// Report all errors
error_reporting(E_ALL);
// Same as error_reporting(E_ALL);
ini_set("error_reporting", E_ALL);
$username_form = $_POST['username'];
$firstname_form = $_POST['firstname'];
$password_form = $_POST['password'];
$you_table_name = 'you_table_name';
$dbfield = array("username"=>$username_form, "firstname"=>$firstname_form,"password"=>$password_form);
$querytest = insertqueryfunction($dbfield,'you_table_name');
if ($conn->query($querytest) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>
用户名：
名字：
密码：
你在哪里使用html表单标签？如果没有这一点，提交表单帖子就无法创建，如果你想使用点击事件，那么就使用AJAXmysql\u real\u escape\u string（）
exists:）@Jack-是的，但它需要一个mysql\u connect
ion，在php的更高版本中，mysql
库被弃用。我显示注释掉的行的原因是，如果他在没有活动连接的情况下尝试我的代码，该行将失败。@Jack-Derp！接得好。：）嗨，山姆，欢迎来到SO，谢谢你注册回答这个问题。为了更好地回答问题，最好添加一些解释，说明代码为什么解决了OP的问题。
<html>
<body>
<form action="dynamicinsert.php" method="POST" >
user name:<br>
<input type="text" id="username" name="username">
<br><br>
first name:<br>
<input type="text" id="firstname" name="firstname">
<br><br>
password:<br>
<input type="password" id="password" name="password">
<br><br>
<input type="submit" name="submit" value="add" />
</form>
</body>
</html>
<?php
$servername = "localhost";
$username = "your_username";
$password = "your_password";
$dbname = "you_DB_name";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
function insertqueryfunction($dbfield,$table) {
$count = 0;
$fields = '';
foreach($dbfield as $col => $val) {
if ($count++ != 0) $fields .= ', ';
$col = addslashes($col);
$val = addslashes($val);
$fields .= "`$col` = '$val'";
}
$query = "INSERT INTO $table SET $fields;";
return $query;
}
if(isset($_POST['submit']))
{
// Report all errors
error_reporting(E_ALL);
// Same as error_reporting(E_ALL);
ini_set("error_reporting", E_ALL);
$username_form = $_POST['username'];
$firstname_form = $_POST['firstname'];
$password_form = $_POST['password'];
$you_table_name = 'you_table_name';
$dbfield = array("username"=>$username_form, "firstname"=>$firstname_form,"password"=>$password_form);
$querytest = insertqueryfunction($dbfield,'you_table_name');
if ($conn->query($querytest) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
?>