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使用HTTP POST方法通过php更新mysql数据库

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它完全在应用程序上执行我仔细检查它 . 它正确发送所有参数，问题是在PHP脚本中它不执行查询（更新，删除）它正确执行插入查询 . php脚本与html表单完美配合，但我不知道问题出在哪里 .

这是我的PHP脚本：

$mysql_host = "localhost";
    $mysql_database = "locator";
    $mysql_user = "root";
    $mysql_password = "";
    mysql_connect($mysql_host,$mysql_user,$mysql_password) or die(mysql_error());
    mysql_select_db($mysql_database);
    $uMail = $_POST['u_mail'];
    $uIMEI = $_POST['u_IMEI'];
    $uName = $_POST['u_name'];
    $uPass = $_POST['u_pass'];
    $tName = $_POST['t_names'];
    $tIMEI = $_POST['t_IMEIs'];
    $tDesc = $_POST['t_desc'];
    mysql_query("UPDATE user_master SET t_names='$tName',t_IMEIs='$tIMEI',t_desc= '$tDesc'      WHERE u_mail = '$uMail' AND u_IMEI = '$uIMEI'");
    mysql_close();
    ?>
php mysql

1 回答

  • 0

    也许你应该用 mysql_real_escape_string() 逃避你的弦乐

    $uMail = mysql_real_escape_string($_POST['u_mail']);
$uIMEI = mysql_real_escape_string($_POST['u_IMEI']);
$uName = mysql_real_escape_string($_POST['u_name']);
$uPass = mysql_real_escape_string($_POST['u_pass']);
$tName = mysql_real_escape_string($_POST['t_names']);
$tIMEI = mysql_real_escape_string($_POST['t_IMEIs']);
$tDesc = mysql_real_escape_string($_POST['t_desc']);
mysql_query("UPDATE user_master SET t_names='$tName',t_IMEIs='$tIMEI',t_desc= '$tDesc'      WHERE u_mail = '$uMail' AND u_IMEI = '$uIMEI'");

    并确保正确设置 $uMail 和 $uIMEI

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