我正在使用jQuery和php执行多个删除记录操作,目前我可以通过点击其工作的复选框删除单个 / 多个记录截至目前很好,但每次删除记录时我的页面都会刷新,因为我没有使用ajax。
我是ajax的初学者我想使用 JQUERY / AJAX 执行相同的操作,每次删除我的记录时都不会重新加载我的页面所以我想使用ajax相同的代码,以便我可以处理我的页面重新加载。
有人帮助我实现它谢谢!!
HTML / PHP
<form method="post" name="data_table">
<table id="table_data">
<tr>
<td>Name</td>
<td>Select All <input type="checkbox" id="check_all" value=""></td>
</tr>
<?php
$query = mysql_query("SELECT * FROM `products`");
while($row = mysql_fetch_array($query))
{
?>
<tr>
<td>
<?php echo $row['product_title']; ?>
</td>
<td>
<input type="checkbox" value="<?php echo $row['id'];?>" name="data[]" id="data">
</td>
</tr>
<?php
}
?>
</table>
<br />
<input name="submit" type="submit" value="Delete" id="submit">
</form>
JQuery的
jQuery(function($)
{
$("form input[id='check_all']").click(function()
{
var inputs = $("form input[type='checkbox']");
for(var i = 0; i < inputs.length; i++)
{
var type = inputs[i].getAttribute("type");
if(type == "checkbox")
{
if(this.checked)
{
inputs[i].checked = true;
}
else
{
inputs[i].checked = false;
}
}
}
});
$("form input[id='submit']").click(function()
{ var inputs = $("form input[type='checkbox']");
var vals=[];
var res;
for(var i = 0; i < inputs.length; i++)
{
var type = inputs[i].getAttribute("type");
if(type == "checkbox")
{
if(inputs[i].id=="data"&&inputs[i].checked){
vals.push(inputs[i].value);
}
}
}
var count_checked = $("[name='data[]']:checked").length;
if(count_checked == 0)
{
alert("Please select a product(s) to delete.");
return false;
}
if(count_checked == 1)
{
res= confirm("Are you sure you want to delete these product?");
}
else
{
res= confirm("Are you sure you want to delete these products?");
}
if(res){
/*** This portion is the ajax/jquery post calling ****/
$.post("delete.php", {data:vals}, function(result){
$("#table_data").html(result);
});
}
});
});
PHP删除代码
<?php
if(isset($_POST['data']))
{
$id_array = $_POST['data']; // return array
$id_count = count($_POST['data']); // count array
for($i=0; $i < $id_count; $i++)
{
$id = $id_array[$i];
$query = mysql_query("DELETE FROM `products` WHERE `id` = '$id'");
if(!$query)
{
die(mysql_error());
}
}?>
答案 0 :(得分:1)
请将更改jquery作为
jQuery(function($)
{
$("form input[id='check_all']").click(function()
{
var inputs = $("form input[type='checkbox']");
for(var i = 0; i < inputs.length; i++)
{
var type = inputs[i].getAttribute("type");
if(type == "checkbox")
{
if(this.checked)
{
inputs[i].checked = true;
}
else
{
inputs[i].checked = false;
}
}
}
});
$("form input[id='submit']").click(function()
{ var inputs = $("form input[type='checkbox']");
var vals=[];
var res;
for(var i = 0; i < inputs.length; i++)
{
var type = inputs[i].getAttribute("type");
if(type == "checkbox")
{
if(inputs[i].id=="data"&&inputs[i].checked){
vals.push(inputs[i].value);
}
}
}
var count_checked = $("[name='data[]']:checked").length;
if(count_checked == 0)
{
alert("Please select a product(s) to delete.");
return false;
}
if(count_checked == 1)
{
res= confirm("Are you sure you want to delete these product?");
}
else
{
res= confirm("Are you sure you want to delete these products?");
}
if(res){
/*** This portion is the ajax/jquery post calling ****/
$.post("delete.php", {data:vals}, function(result){
$("#table_data").html(result);
});
}
});
});
删除.php为
<?php
if(isset($_POST['data']))
{
$id_array = $_POST['data']; // return array
$id_count = count($_POST['data']); // count array
for($i=0; $i < $id_count; $i++)
{
$id = $id_array[$i];
$query = mysql_query("DELETE FROM `test` WHERE `id` = '$id'");
if(!$query)
{
die(mysql_error());
}
}?>
<tr>
<td>ID</td>
<td>TITLE</td>
<td>Select All <input type="checkbox" id="check_all" value=""></td>
</tr>
<?php
$query = mysql_query("SELECT * FROM `test`");
while($row = mysql_fetch_array($query))
{
?>
<tr>
<td>
<?php echo $row['id']; ?>
</td>
<td>
<?php echo $row['name']; ?>
</td>
<td>
<input type="checkbox" value="<?php echo $row['id'];?>" name="data[]" id="data">
</td>
</tr>
<?php
} unset($row);
}