I am using php,mysql and ajax to delete record from a table. The problem is that the in MySQL_query it not getting the id it shows "id= undefined", i tried to pass the id to the query but i don't know where i went wrong i tried to print MySQL its shows
我使用php、mysql和ajax从表中删除记录。问题是,在MySQL_query中它没有获得id它显示为"id= undefined"我试图将id传递给查询但我不知道哪里出错了我试图打印MySQL显示
delete from 9xx WHERE id = undefinedArray
(
[rowid] => undefined
[supplier] => 9xx
)
can anyone tell me how to pass the id ...thanks
谁能告诉我怎么通过身份证吗
My ajax
我的ajax
$(".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');
}});
}
});
My html
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
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($_POST);
?>
2 个解决方案
#1
2
<td style="width:62px" class="deletesuppliernetwork '.$rows["id"].'"><img src="/image/delete.png" style="margin:0 0 0 17px" ></td>
the index of your ID is 1, that is second index. not 2.
ID的索引是1,这是第二个索引。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');
}});
#2
1
var rowObj = $(this);
$.ajax({
type: "POST",
url: "suppliernetwork/delete.php",
data: "rowid="+arr[1]+"&supplier="+supplier,
success: function(data){
$('.ajax').html($('.ajax input').val());
$('.ajax').removeClass('ajax');
$(rowObj).parents("tr:first").hide();
}});
this should hide your complete row.
这将隐藏您的完整行。
#1
2
<td style="width:62px" class="deletesuppliernetwork '.$rows["id"].'"><img src="/image/delete.png" style="margin:0 0 0 17px" ></td>
the index of your ID is 1, that is second index. not 2.
ID的索引是1,这是第二个索引。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');
}});
#2
1
var rowObj = $(this);
$.ajax({
type: "POST",
url: "suppliernetwork/delete.php",
data: "rowid="+arr[1]+"&supplier="+supplier,
success: function(data){
$('.ajax').html($('.ajax input').val());
$('.ajax').removeClass('ajax');
$(rowObj).parents("tr:first").hide();
}});
this should hide your complete row.
这将隐藏您的完整行。