Question

我已经尝试搞清楚错误很长一段时间了。即使尝试使用尝试和捕获但无济于事。我得到的输出是“查询失败”（这是其他部分）。代码也可以在不同的机器上运行。任何人都可以帮我解决这个问题吗？

<?php
session_start();

//Include database connection details
    require_once('connect.php');

    //Array to store validation errors
    $errmsg_arr = array();

    //Validation error flag
    $errflag = false;

//Function to sanitize values received from the form. Prevents SQL injection

    function clean($str) {
    $str = @trim($str);
    if(get_magic_quotes_gpc()) {
        $str = stripslashes($str);
    }
    return mysql_real_escape_string($str);
}
    try {
    //Sanitize the POST values
    $username = clean($_POST['username']);
    $phone = clean($_POST['phone']);
    /*$username = 'Ajay M';
    $phone = '9810827804';*/
    //Input Validations
    if($username == '') {
        $errmsg_arr[] = 'Username missing';
        $errflag = true;
    }
    if($phone == '') {
        $errmsg_arr[] = 'phone missing';
        $errflag = true;
    }

//If there are input validations, redirect back to the login form
if($errflag) {
    $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
    session_write_close();
    header("location: log.php");
    exit();
}
//Create query

        $qry="SELECT * FROM app_user WHERE name='$username' AND phone='$phone'";
    $result=mysql_query($qry);
    //Check whether the query was successful or not
    if($result) {
        if(mysql_num_rows($result) > 0) {
            //Login Successful
            session_regenerate_id();
            $member = mysql_fetch_assoc($result);
            $_SESSION['SESS_MEMBER_ID'] = $member['id'];
            $_SESSION['SESS_NAME'] = $member['username'];
            $_SESSION['SESS_PHONE'] = $member['phone'];
            session_write_close();
            header("location: suzuki_home.php");
            exit();
        }else {
            //Login failed
            $errmsg_arr[] = 'user name and phone not found';
            $errflag = true;
            if($errflag) {
                $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
                session_write_close();
                header("location: fail.php");
                exit();
            }
        }
    }else {
        die("Query failed");
    }
}
catch (Exception $e) {
    echo ($e->getMessage());
}
?>

Answer 1

name 是MySQL上的保留字。因此，请在查询中使用反引号括起列名。

$qry="SELECT * FROM `app_user` WHERE `name`='$username' AND `phone`='$phone'";

Answer 2

必须在.ini文件中设置错误报告，但您也可以在.php文件中执行此操作：

ini_set("display_errors", true);
error_reporting(E_ALL);

PHP代码不工作，没有提到错误

2 个答案: