jquery ajax更新mysql表

在我的程序中有mysql表'company'.companyid和activation_code是該表中的一些列.i顯示錶中的值在html表中包含comapproval.php page.that html表每行都有一個接受按鈕，當我點擊該按鈕時，公司表應該更新，該行從html table.update中刪除是'approvecompany.php'page.i使用jquery ajax for this.when我cliick接受按鈕，我得到只成功alert.but我不能做updaet table.even sucessfull警報只獲得第一行如果html table.i是新的ajax.help我來解決這個問題。jquery ajax更新mysql表

comapproval.php

<table class="table table-striped table-bordered table-list"> 
        <thead> 

        <tr> 
         <th>Action</th> 
         <th>ID</th> 
         <th>Registration number</th> 
         <th>Company Name</th> 
         <th>Email</th> 
        </tr> 
        </thead> 
        <tbody> 

        <?php 
        $conn = mysqli_connect("localhost", "root", "", "internship"); 
        if (mysqli_connect_errno()) { 
         echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
        } 
        $sql = ""; 

        $sql = "select * from company where activation_code=0"; 

        $res = mysqli_query($conn, $sql); 

        while ($row = mysqli_fetch_assoc($res)): 

        ?> 

        <tr> 
         <td align="center"><input type="submit" class="btn btn-default" value="Accept" id="accept" name="accept"></input></td> 
         <td><?php echo $row['companyid']; ?></td> 
         <td><?php echo $row['government_reg_no']; ?></td> 
         <td><?php echo $row['company_name']; ?></td> 
         <td><?php echo $row['email']; ?></td> 
        </tr> 

        <?php 
         endwhile 
        ?> 

        </tbody> 
       </table> 
<script> 
      $("#accept").click(function() { 
       $.ajax({ 
        type:"POST", 
        url:"approvecompany.php", 
        data:{comid:$('<?php $row['companyid']?>').val()}, 
        success:function() { 
         alert('Successfully approved'); 
         window.location.reload(true); 
        } 

       }); 
      }); 

    </script>

approvecompany.php

<?php 
$conn=mysqli_connect("localhost","root","","internship"); 

$comid=$_POST['comid']; 

if (mysqli_connect_errno()) { 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$sql="update company set activation_code=1 where companyid=$comid"; 

if ($conn->query($sql) === TRUE) { 

?> 

<?php 
} 

else{ 
?><script>alert("Error...")</script><?php 
} 
?>

來源

2017-04-13 SRLA

[小巴比]（HTTP：//巴比-tables.com/）說* ** [您的腳本存在SQL注入攻擊風險。]（http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php）***瞭解[已準備好]（http://en.wikipedia.org/wiki/Prepared_statement）對[MySQLi]的聲明（http://php.net/manual/en/mysqli.quickstart.prepared-statements.php）。即使[轉義字符串]（http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string）是不安全的！ –

[你看過瀏覽器開發工具中的AJAX請求/響應了嗎？你有沒有在項目中包含jQuery庫？是否有任何錯誤報告？你是在網絡服務器上運行這個嗎？]（http://jayblanchard.net/basics_of_jquery_ajax.html） –

[ID必須是唯一的]（http://stackoverflow.com/questions/5611963/can-multiple-different- html-elements-have-the-same-id-if-theyre-different-eleme），具體是因爲它會導致[JavaScript]中的問題（https://developer.mozilla.org/en-US/docs/Web/ HTML/Global_attributes/id）和CSS，當你嘗試與這些元素進行交互時。如果你有多行'id ='accept''，事情就不會按計劃進行。將其更改爲一個類。 –

嘗試如下重寫comapproval.php：

<table class="table table-striped table-bordered table-list"> 
    <thead> 
     <tr> 
      <th>Action</th> 
      <th>ID</th> 
      <th>Registration number</th> 
      <th>Company Name</th> 
      <th>Email</th> 
     </tr> 
    </thead> 
    <tbody> 

    <?php 
     $conn = mysqli_connect("localhost", "root", "", "internship"); 
     if (mysqli_connect_errno()) { 
      echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
     } 
     $sql = ""; 

     $sql = "select * from company where activation_code=0"; 

     $res = mysqli_query($conn, $sql); 

     while ($row = mysqli_fetch_assoc($res)): 

    ?> 

     <tr> 
      <td align="center"> 
       <button type="submit" class="btn btn-default myButton" value="<?php echo $row['companyid']; ?>" id="accept" name="accept">Accept</button> 
      </td> 
      <td><?php echo $row['companyid']; ?></td> 
      <td><?php echo $row['government_reg_no']; ?></td> 
      <td><?php echo $row['company_name']; ?></td> 
      <td><?php echo $row['email']; ?></td> 
     </tr> 

    <?php 
     endwhile 
    ?> 

    </tbody> 
</table> 
<script> 
    $(".myButton").click(function() { 
     var company_id = $(this).val(); 
     $.ajax({ 
      type:"POST", 
      url:"approvecompany.php", 
      data:{ comid: company_id }, 
      success:function() { 
       window.location.reload(true); 
      } 

     }); 
    }); 
</script>

來源

2017-04-13 20:31:30

100％works.thank u非常適合你貢獻薩欽PATIL ....你是一個偉大的朋友...... – SRLA

jquery ajax更新mysql表

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