Ajax插入数据库中的错误 [英] Error in ajax insert database

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本文介绍了Ajax插入数据库中的错误的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

在每一行中,我想添加一个可编辑的td来将"mavandon"插入DATABASE dsdonhang中.我在 http://phppot.com/php/上阅读了此内容php-mysql-inline-editing-using-jquery-ajax/,但不起作用:((

In each row, i want to add a editable td to insert "mavandon" into DATABASE dsdonhang. I read this on http://phppot.com/php/php-mysql-inline-editing-using-jquery-ajax/ but it doesn't work :((

<script src="../lib/jquery-3.1.0.min.js"></script>   
<script>
$(document).ready(function(){        
    function saveToDatabase(editableObj,column,idd) {
        $.ajax({
            url: "saveedit.php",
            type: "POST",
data:'column='+$(this).column+'&editval='+$(this).editableObj.innerHTML+'&idd='+$(this).idd,
            success: function(data){
                $(editableObj).css("background","#FDFDFD");
                //alert ("hello");
            }
        error: function() {}    
       });
    }
});        
</script>
<?php $query = mysqli_query($conn,"SELECT * FROM dsdonhang");            
    while($row=mysqli_fetch_assoc($query)) {$data[] = $row;}
    foreach($data as $k=>$v) {
      ?>
        <tr>
          <td><?php $madon = $data[$k]["idd"];echo $k+1; ?></td>
          <td><?php echo $data[$k]["ngaydat"]; ?></td>
          <td><?php echo $data[$k]["hoten"]; ?></td>
          <td><?php echo $data[$k]["diachi"]; ?></td>
          <td><?php echo $data[$k]["sdt1"]; ?></td>
          <td><?php echo $data[$k]["donhang"]; ?></td>
          <td><?php echo $data[$k]["tongtien"]; ?>.000VN&#272;</td>
          <td><?php echo $data[$k]["nguoinhan"]; ?></td>
          <td contenteditable="true" onchange="saveToDatabase(this,'mavandon','<?php echo $madon; ?>')"><?php echo $data[$k]["mavandon"]; ?></td>
          <td><a href="donhang.php?id=<?php echo $data[$k]["idd"]; ?>&tc=<?php echo $data[$k]["sothamchieu"]; ?>">Xem</a></td>
        </tr>
<?php
  }            
?>

saveedit.php

saveedit.php

<?php
require_once("../lib/connection.php");
mysqli_query($conn,"UPDATE dsdonhang set " . $_POST["column"] . " = '".$_POST["editval"]."' WHERE  idd='".$_POST["idd"]."'");
?>

有帮助吗?谢谢

推荐答案

尝试一下:

<script>
$(document).ready(function(){        
function saveToDatabase(editableObj,column,idd) {
     var data = {'column':$(this).column,'editval':$(editableObj).text(),'idd':$(this).idd };
    $.ajax({
        url: "saveedit.php",
        type: "POST",
        contentType: "application/json; charset=utf-8",
        datatType: "json",
        data:JSON.stringify(data),
        success: function(data){
            $(editableObj).css("background","#FDFDFD");
            console.log("received data=>"+data);
            alert ("hello");
        }
    error: function(err) {
       console.log("error=>"+err); //print error if exist
    }    
   });
}
});        
</script>

这篇关于Ajax插入数据库中的错误的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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