I'm using this jQuery code:
我使用的是jQuery代码:
$.ajax
({
type: "POST",
url: "customerfilter.php",
data: dataString,
cache: false,
success: function(html)
{
$(".custName").html(html);
}
});
How can i do something like this: $(".projDesc").html(html1);
So i can split the returned results into two html elements?
我如何做这样的事情:$(".projDesc").html(html1);所以我可以把返回的结果分成两个html元素?
echo "<p>" .$row['cust_name']. "</p>";
thats the PHP i'm using and i want to echo another statement which i can put into another HTML element
这就是我正在使用的PHP,我想要回显另一个语句,我可以将它放入另一个HTML元素中
Does this make sense?
这是否有意义吗?
3 个解决方案
#1
55
Use json_encode()
to convert an associative array from PHP into JSON and use $.getJSON()
, which will return a Javascript array.
使用json_encode()将关联数组从PHP转换为JSON,并使用$. getjson(),它将返回一个Javascript数组。
Example:
例子:
<?php echo json_encode(array("a" => "valueA", "b" => "valueB")); ?>
In Javascript:
在Javascript中:
$.getJSON("myscript.php", function(data) {
alert("Value for 'a': " + data.a + "\nValue for 'b': " + data.b);
});
#2
42
Make your response return JSON, you'll need to change your jQuery to this, so the expected dataType is json:
让您的响应返回JSON,您需要将jQuery更改为这个,因此期望的数据类型是JSON:
$.ajax
({
type: "POST",
url: "customerfilter.php",
dataType: 'json',
cache: false,
success: function(data)
{
$(".custName").html(data.message1);
$(".custName2").html(data.message2);
}
});
Then you need to encode your response as a JSON Array:
然后需要将响应编码为JSON数组:
<?php echo json_encode(
array("message1" => "Hi",
"message2" => "Something else")
) ?>
#3
3
Why don't you return a JSON object. This way you can easily put many different results inside the ajax response.
为什么不返回一个JSON对象呢?通过这种方式,您可以很容易地将许多不同的结果放入ajax响应中。
#1
55
Use json_encode()
to convert an associative array from PHP into JSON and use $.getJSON()
, which will return a Javascript array.
使用json_encode()将关联数组从PHP转换为JSON,并使用$. getjson(),它将返回一个Javascript数组。
Example:
例子:
<?php echo json_encode(array("a" => "valueA", "b" => "valueB")); ?>
In Javascript:
在Javascript中:
$.getJSON("myscript.php", function(data) {
alert("Value for 'a': " + data.a + "\nValue for 'b': " + data.b);
});
#2
42
Make your response return JSON, you'll need to change your jQuery to this, so the expected dataType is json:
让您的响应返回JSON,您需要将jQuery更改为这个,因此期望的数据类型是JSON:
$.ajax
({
type: "POST",
url: "customerfilter.php",
dataType: 'json',
cache: false,
success: function(data)
{
$(".custName").html(data.message1);
$(".custName2").html(data.message2);
}
});
Then you need to encode your response as a JSON Array:
然后需要将响应编码为JSON数组:
<?php echo json_encode(
array("message1" => "Hi",
"message2" => "Something else")
) ?>
#3
3
Why don't you return a JSON object. This way you can easily put many different results inside the ajax response.
为什么不返回一个JSON对象呢?通过这种方式,您可以很容易地将许多不同的结果放入ajax响应中。