Php 表单不更新数据库
我有一个从员工列表中填充的更新表单。数据库中的值正在传递,但未更新。这是我的代码和我正在显示的内容Php 表单不更新数据库,php,sql,database,Php,Sql,Database,我有一个从员工列表中填充的更新表单。数据库中的值正在传递,但未更新。这是我的代码和我正在显示的内容 <?php $con = mysql_connect("localhost","root","*******"); if (!$con) { die('Could not connect: ' . mysql_error()); } $query = mysql_query("select * from backup"); if(isset($_POST['update
<?php
$con = mysql_connect("localhost","root","*******");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
$query = mysql_query("select * from backup");
if(isset($_POST['update']))
$id = $_POST['id'];
$first = $_POST['first'];
$last = $_POST['last'];
$store = $_POST['store'];
$title = $_POST['title'];
$title2 = $_POST['other'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$dept = $_POST['dept'];
$bio = $_POST['bio'];
$query="UPDATE backup SET first='$first', last='$last', store='$store', title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
mysql_query($query);
echo "Record Updated";
mysql_close();
print_r($_POST)
?>
您还应该选择数据库:
// make foo the current db
$db_selected = mysql_select_db('foo', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
但最重要的是,您应该使用MySQLi或PDO_-MySQL。这是因为,您忘记了{
在if(isset(…)
之后
此外,未选择任何数据库
更正代码如下:
<?php
$con = mysql_connect("localhost","root","*******");
mysql_select_db('DB_NAME');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
$query = mysql_query("select * from backup");
if(isset($_POST['update'])) {
$id = $_POST['id'];
$first = $_POST['first'];
$last = $_POST['last'];
$store = $_POST['store'];
$title = $_POST['title'];
$title2 = $_POST['other'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$dept = $_POST['dept'];
$bio = $_POST['bio'];
$query="UPDATE backup SET first='$first', last='$last', store='$store', title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
mysql_query($query);
echo "Record Updated";
mysql_close();
print_r($_POST)
}
?>
您尚未在连接中定义数据库
您的代码可以更正为:
<?php
$con = mysqli_connect("localhost","root","**","db");
if (!$con)
{
die('Could not connect: ' . mysqli_error());
}
//$query = mysqli_query($con,"select * from backup");
if(isset($_POST['update'])):
$id = $_POST['id'];
$first = $_POST['first'];
//other stuffs
$query="UPDATE backup SET first='$first', last='$last', store='$store', title='$title', title2='$title2', phone='$phone', email='$email', bio='$bio' WHERE id='$id'";
$res = mysqli_query($con,$query);
if(!$res)
die("could not update records. Error = ".mysqli_error($con));
echo "Record Updated";
mysqli_close($con);
print_r($_POST);
endif;
?>
不再支持mysql\uquery*
函数,它们已不再维护,将来也将保留。您应该使用或更新代码,以确保将来项目的功能。您可以用mysql\u query($query)替换mysql\u query,还是用mysql\u query($query)替换mysql\u query(mysql\u error())因此,您可以查看查询是否有任何错误?您可能还希望将此部分:if(isset($\u POST['update'])用大括号{}括起来谢谢Neal.我会带着这个去MySQLi.谢谢你.我应该而且会改正的.我又回到了一个旧习惯谢谢你.我一定是把它删除了.因为它原来就在那里:)谢谢…我本来就有数据库.这可能就是我错过它的原因.还有,花括号又杀了我.你帮了大忙!!谢谢你额外的一双眼睛..显然我的眼睛不好:)