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php-如何使用Ajax从数据库中删除记录

2019-10-12 19:20:06  阅读:284  来源: 互联网

标签:php jquery sql mysql ajax



我正在使用php,mysql和ajax从表中删除记录.问题是,在MySQL_query中,它没有获取显示为“ id = undefined”的ID,我试图将ID传递给查询,但是我不知道我哪里出了错,我试图打印MySQL的节目

delete from 9xx WHERE id = undefinedArray
(
    [rowid] => undefined
    [supplier] => 9xx
)

谁能告诉我如何通过身份证…谢谢

我的阿贾克斯

$(".deletesuppliernetwork").live('click',function()
        {
         arr = $(this).attr('class').split( " " );
        var supplier=document.getElementById("supplier").value;

        if(confirm("Sure you want to delete this update?"))
        {
        $.ajax({
        type: "POST",
        url: "suppliernetwork/delete.php",
        data: "rowid="+arr[2]+"&supplier="+supplier,
        success: function(data){
                                                         $('.ajax').html($('.ajax input').val());
                                                         $('.ajax').removeClass('ajax');
                                                    }});
        }
        });

我的HTML

<?php
include"db.php";

$supplier_id=$_GET['supplier_id'];

if($supplier_id!=""){

$sql=mysql_query("select * from $supplier_id order by country,networkname" );

while($rows=mysql_fetch_array($sql))
{

if($alt == 1)
        {
           echo '<tr class="alt">';
           $alt = 0;
        }
        else
        {
           echo '<tr>';
           $alt = 1;
        }

echo '  <td style="width:123px" class="edit supplier '.$rows["id"].'">'.$rows["supplier"].'</td>
                <td style="width:104px" class="edit rn '.$rows["id"].'">'.$rows["rn"].'</td>    
            <td style="width:103px" class="edit sc '.$rows["id"].'">'.$rows["sc"].'</td>    
            <td style="width:108px" class="edit comment '.$rows["id"].'">'.$rows["comment"].'</td>

            <td style="width:62px" class="deletesuppliernetwork '.$rows["id"].'"><img   src="/image/delete.png" style="margin:0 0 0 17px" ></td>                                

        </tr>';


}
}
?>

delete.php

<?php
    include"db.php";

$supplier=$_POST['supplier'];




        $rownum=$_POST['rowid'];  
        $sql="delete from $supplier WHERE id = ".$rownum."";

        print $sql;

        mysql_query($sql);  


    print_r($_POST);
?>

解决方法:

<td style="width:62px" class="deletesuppliernetwork '.$rows["id"].'"><img   src="/image/delete.png" style="margin:0 0 0 17px" ></td>     

您ID的索引是1,即第二个索引.不是2.

$.ajax({
        type: "POST",
        url: "suppliernetwork/delete.php",
        data: "rowid="+arr[1]+"&supplier="+supplier,
        success: function(data){
                                                         $('.ajax').html($('.ajax input').val());
                                                         $('.ajax').removeClass('ajax');
                                                    }});



标签:php,jquery,sql,mysql,ajax

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