dongza3124 2017-07-08 10:56
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用户名不从数据库中获取

I'm new to php. I've some problem. I get the value of username with ($_GET) variable. Let's suppose if some one enter (mehwish) in url. I want to check if the username entered by user is available or not in the database. My code below doesn't work. Correct my mistake if there's any. Thank you for your support.

<?php include("header.php");
    include("classes/db.php");
    session_start();
    $username="";
    if(isset($_GET['username']))
    {
        $user=$_GET['username'];
        if("SELECT user_name FROM user_reg WHERE user_name='$user'")
        {
            $username="SELECT user_name FROM user_reg WHERE user_name='$user'";
            $run_username=mysqli_query($con,$username);
            $row=mysqli_fetch_array($run_username);
            $username=$row['user_name'];
        }
        else
        {
            die("User does nto exist");
        }
    }
    ?>
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  • dongzhuohan7085 2017-07-08 11:01
    关注

    Use something like this :

    <?php 
    session_start(); //This always need to be first line of code
    include("header.php");
    include("classes/db.php");
    
    $username="";
    if(isset($_GET['username']))
    {
        $user = mysqli_real_escape_string($con,$_GET['username']);
        //get data from table using the passed user name in the url
        $checkUser = mysqli_query($con,"SELECT user_name FROM user_reg WHERE user_name='$user'") or die(mysqli_error($con));
        //return number of rows from the above query
        //if more than 0 means user exist in database table
        if(mysqli_num_rows($checkUser)>0)
        {
            $row = mysqli_fetch_array($checkUser);
            $username = $row['user_name'];
            //use like this to store in session using $_SESSION global variable
            $_SESSION["firstname"] = $row["firstname"]; //firstname - column name in table
            $_SESSION["lastname"] = $row["lastname"];
            $_SESSION["email"] = $row["email"];
        }
        else
        {
            echo "User does not exist";
        }
    }
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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