PHP脚本不返回查询，但MySql查询工作

我的最终目标是让一个js脚本以json的形式返回一个PHP查询的结果，这样我就可以用它做出恶意的事情。PHP脚本不返回查询，但MySql查询工作

我有我想要使用的MySql查询，当我在Workbench中测试它时它肯定正在工作，但是当我在PHP脚本中进行测试时，没有任何东西会返回。 PHP中的当前查询只是测试数据交换的占位符。

我手动把一些东西放到数组（searchResults）并且得到返回，但没有任何东西来自PHP脚本执行时。

我也明白这可能不是最安全或最有效的代码，我只是想在这一点上得到它的工作。

<?php 
    // Include your database creds and login to the db 
    require_once 'login_karavites.php'; 
    $db = mysqli_connect($db_hostname, $db_username, $db_password); 

    // Handle the input/request. 
    $searchString_UNSAFE = $_POST['eName']; // change that, obviously 

    // Bare minimum sanitation to prevent injection. 
    $searchString = $db->escape_string($searchString_UNSAFE); 

    // Construct the SQL query 
    $sql = "SELECT * FROM `Halls` WHERE hall_name = 'Rose Ballroom'"; 

    // Do the database lookup. 
    $result = $db->query($sql); 

    // Create empty array to hold our results (to be sent back to the browser). 
    $searchResults = array(); 
    $searchResults[]="wow"; 
    // If we had results, put them into that array 
    if ($result->num_rows > 0) { 

     // This loop will retrieve every row from that result set 
     while ($row = $result->fetch_assoc()) { 

      // From each row, just take the 'event_name' field. 
      $searchResults[] = $row['hall_name']; 

     } 

    } 

    // Done with the db, now we just have to send the results back to the browser. 
    $db->close(); 

    // Send the correct content-type header. 
    // This ensures that jQuery automatically converts the response into an 
    // array or object, rather than just treating it like a block of text. 
    // Must be the FIRST thing the PHP script outputs, or it will choke. 
    header('Content-type: application/json'); 

    // Output the data. 
    echo json_encode($searchResults); 

?>

该js脚本。

$(document).ready(function() { 
    // All this stuff runs as soon as the page is fully loaded 

    // Attach a function to the Submit action on #eventForm 
    $('#eventForm').submit(function() { 

     // Submit the form via AJAX 
     $(this).ajaxSubmit({ 

      // Attach a function to the "the PHP script returned some results" event 
      success: function(response, status, xhr, $form){ 
       // I am assuming that this is your data format, for example: 
       // { "searchResults": [ "result1", "result2", "result3" ] } 
       // I am also assuming that you want your results in div#results 
       $('div#results').html(""); // Clear it out of anything that's already there. 
       console.log(response); 
       for (i in response['searchResults']) { 
        $('div#results').append(response['searchResults'][i]); 
       } 
      }, 

      // Give up if PHP doesn't answer in 3 seconds 
      timeout: 3000, 

      // Path to the PHP file we want to send this to 
      url: 'phpdata/eventsData.php' 
     }); 

     // Make sure the browser does NOT proceed to submit the form again, 
     // the old fashioned way (full page reload). 
     return false; 
    }); 
});

来源

2014-05-05 Zeratas

您是否尝试使用浏览器直接调用您的php脚本？你能看到什么吗？如果答案是'是'和'否'，您是否尝试在设置标题之前转储$ searchResults的内容？ –

当我尝试直接调用它时，我只能得到那一行。如果我注释掉setHeaders行，这也是同样的事情。 – Zeratas

你有一个未使用的变量'$ searchString' – sectus

哪一部分我的问题之一是：

我愣神从未设置形式，方法在原来的形式，因此POST实际上并没有经历。
我在PHP文件中设置了错误的MySql连接。

我自己的笔记，请确保您检查所有的设置。

来源

2014-05-05 14:55:47 Zeratas

PHP脚本不返回查询，但MySql查询工作

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