我有以下脚本，它应该产生错误，但它没有:
<?php
error_reporting( E_ALL | E_STRICT );
ini_set('display_errors', 1);
?>
<html>
<head>
<title></title>
<link rel="icon" type="image/png" href="favicon.ico">
<?php
$err = array();
if( $_SERVER['REQUEST_METHOD']=='POST' ) {
if( empty( $_POST['display_name'] ) ) $err[] = "display name field is required";
if( empty( $_POST['email'] ) ) $err[] = "email field is required";
if( empty( $_POST['password'] ) ) $err[] = "password field is required";
if( !$err ) {
try {
$DBH = new PDO( "mysql:host=localhost;dbname=database1", "user", "pass" );
$DBH -> setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
$STH = $DBH -> prepare( "delete into table1 (display_name, email, password) values ( :display_name, :email, :password )" );
$STH -> bindParam( ':display_name', $_POST['display_name'], PDO::PARAM_STR, 100 );
$STH -> bindParam( ':email', $_POST['email'], PDO::PARAM_STR, 100 );
$STH -> bindParam( ':password', $_POST['password'], PDO::PARAM_STR, 100 );
$STH -> execute();
$STH = $DBH -> prepare( "delete into table2 ( username, status, users_id ) values ( :username, :status, :users_id )" );
$strStatus = 1;
$STH -> bindParam( ':username', $_POST['display_name'], PDO::PARAM_STR, 100 );
$STH -> bindParam( ':status', $strStatus, PDO::PARAM_INT, 1 );
$STH -> bindParam( ':users_id', $_POST['referer'], PDO::PARAM_INT, 1 );
$STH -> execute();
$DBH = null;
} catch( PDOException $e ) {
echo $e -> getMessage();
}
header( "Location: ".$_SERVER['PHP_SELF'] );
exit;
} else {
foreach( $_POST as $key => $val ) {
$form[$key] = htmlspecialchars($val);
}
}
} else {
$form['display_name'] = $form['email'] = $form['password'] = '';
}
?>
</head>
<body>
<?php foreach( $err as $line ) { ?>
<div style="error"><?php echo $line; ?></div>
<?php } ?>
<h1>register</h1>
<form method="post">
referers id:<br />
<input type="text" name="referer" /><br /><br />
name:<br />
<input type="text" name="display_name" value="<?php echo $form['display_name']; ?>" /><br /><br />
email:<br />
<input type="text" name="email" value="<?php echo $form['email']; ?>" /><br /><br />
password:<br />
<input type="text" name="password" value="<?php echo $form['password']; ?>" /><br /><br />
<input type="submit" value="register" />
</form><br /><br />
or register with one of your existing accounts:<br /><br />
facebook<br /><br />
google
</body>
</html>
查看 2 个 SQL 查询...
最佳答案
我不认为 execute() 会引发异常(如果我错了，请有人纠正我)。
它将准备好的语句(查询)发送到数据库并根据成功返回 TRUE 或 FALSE。
您可以检查返回值并回显 $STH -> error 以查看 MySQL 错误消息。像这样:
$success = $STH -> execute();
if (! $success)
echo $STH -> error
关于php - 删除到 table1 不产生错误，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/5438551/