javascript 更改ajax调用复选框
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checkbox on change ajax call
提问by Muhammad Atiq
I have one checkbox. I want when I checked the checkbox so I get the 1 and then update a mysql query through that get value. I also want if I unchecked the checkbox so I get a value 0 so then I again update the mysql query. Help me. it should be done with ajax call. code will be in PHP.
我有一个复选框。当我选中复选框时，我希望得到 1，然后通过该获取值更新 mysql 查询。如果我取消选中该复选框，我还想得到一个值 0，然后我再次更新 mysql 查询。帮我。它应该通过ajax调用来完成。代码将在 PHP 中。
HTML code
HTML代码
<input type="checkbox" name="action1" id="action1" title="Action 1" value="1" onclick="return Populat_Industry('set_home_vid.php');"/>
ajax call
ajax调用
<script>
function Populat_Industry(url){
var value=$(#action1).val();
$.ajax({
type: "POST",
url: url,
async: true,
data: "value="+value,
success: function(msg){
//alert('Success');
if(msg !='success'){
//alert('Fail');
}
}
});
}
</script>
PHP code
PHP代码
if($_POST['action1']=='1'){
$query= mysql_query("UPDATE homevideos SET is_active = '1
}
else{
mysql_query("UPDATE homevideos SET is_active = '0')
echo 'success';
回答by Ehsan Sajjad
Ajax call is async. you cannot use returnwith it this way. Write checkbox change event in jquery and send ajax call.
Ajax 调用是异步的。您不能return以这种方式使用它。在 jquery 中编写复选框更改事件并发送 ajax 调用。
Do like this:
这样做：
HTML:
HTML：
<input type="checkbox" name="action1" id="action1" title="Action 1" value="1"/>
JQUERY:
查询：
$("#action1").change(function () {
var value = $(this).val();
$.ajax({
type: "POST",
url: "set_home_vid.php",
async: true,
data: {
action1: value // as you are getting in php $_POST['action1']
},
success: function (msg) {
alert('Success');
if (msg != 'success') {
alert('Fail');
}
}
});
});
回答by Sharad Kale
HTML :
HTML :
<input type="checkbox" name="action1" id="action1" title="Action 1" value="1" url="set_home_vid.php" />
JQuery:
查询：
<script>
$("#action1").change(function(){
var value = $(this).val();
var url = $(this).attr("url");
$.ajax({
type: "POST",
url: url,
data: "value="+value, //POST variable name value
success: function(msg){
if(msg =='success'){
alert('Success');
}
else{
alert('Fail');
}
}
});
});
</script>
PHP:
PHP：
if($_POST['value']==1){ //as used variable name "value" in ajax post data
$query= mysql_query("UPDATE homevideos SET is_active = 1"); //query was incomplete and missing ";"
echo 'success';
}
else{
mysql_query("UPDATE homevideos SET is_active = 0); // missing ";"
echo 'success';
}
回答by MikeVelazco
Maybe you could use the .isjquery method
也许你可以使用.isjquery 方法
something like this:
像这样：
$("#i").bind("change",function(){
if($(this).is(":checked"))
// set value for ajax
else
// set another value for ajax
// ajax code here
});
$("#i").bind("change",function(){
if($(this).is(":checked"))
// set value for ajax
else
// set another value for ajax
// ajax code here
});
回答by Brian Bolli
You forgot to put quotes around your call to get the input value:
您忘记在您的电话周围加上引号以获取输入值：
var value=$("#action1").val()
