php - 在mysql中执行两个php变量where查询

Question

我的 html 表单 ->

<form action="where.php" method="post">
    <h2 align="center" style="color: white;"> Search Challan </h2>
    <table border="1"  bgcolor="grey" align="center">   
        <tr>
            <td align="center"> Search a Challan Details . Enter the Challan no below :</td>
        </tr>
        <tr>
            <td align="center">
        </tr>
        <tr>
            <td align="center"><input type="submit" name="submit" value="Search"align="middle" ></td> 
        <tr>
            <td>
                <select name="squery" style="width:142px;" >
                    <option value="challan_no">Challan no </option>
                    <option value="product_name">Product Name</option>
                    <option value="Buyer">buyer</option>
                    <option value="Employee Responsible">Employee</option>
            </td>
        </tr>
        <tr>
            <td>
                <input type="text" name="search" >
            </td>       
        </tr>      
    </tr>
</form>

where.php ->

<?php
    $dbhost='localhost';
    $dbusername='root';
    $dbuserpass='';
    $dbname='inventory';

    //connect to the mysql database server.
    $con = mysql_connect ($dbhost, $dbusername, $dbuserpass);

    if (!$con ) die ("unable to connect : ". mysql_error());

    mysql_selectdb("$dbname",$con ) ; 

    $user_req = $_REQUEST['squery'] ;     //colomn name
    $req_id = $_REQUEST['search'] ;       // 

    $query = "SELECT * FROM challan WHERE '$user_req' = $req_id  ";
    $result = mysql_query($query);  
    if (!$result) die ("DAtabase acces faild bc : ". mysql_error());       

    $rows = mysql_numrows($result);   

    for ($j=0 ; $j < $rows ; ++$j)
    {
        $row = mysql_fetch_row($result);

        echo "<TABLE border=1 bgcolor=grey align=center width=500px float=left>"   ;
        echo "<tr>";
        echo "<td align=center>Challan no :  </td>";
        echo " <td> $row[0]  </td>";
        echo " </tr>";
        echo "<tr>";    
        echo "<td align=center>Challan Date : </td>"; 
        echo " <td> $row[1]  </td>";
        echo "<tr>";    
        echo "<td align=center>Product Name :   </td> ";
        echo " <td> $row[2]  </td>";
        echo "<tr>";    
        echo "<td align=center>Product qty :  </td> " ;
        echo " <td> $row[3]  </td>";
        echo "<tr>";    
        echo "<td align=center> Buyer : </td> " ;
        echo " <td> $row[4]  </td>";
        echo "<tr>";    
        echo "<td align=center>Employee Responsible :  </td> " ;
        echo " <td> $row[5]  </td>";
        echo "</tr>"; 
    }
?>

它的输出只是空白。我无法确定我哪里出了问题。

Answer 1

空白屏幕表示发生了 fatal error ，并且 php 无法继续执行脚本，您可以在脚本顶部运行此命令，以将错误作为输出的一部分打印到屏幕上。

<?php
ini_set('display_errors', 1);