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Home Backend Development PHP Tutorial 利用ajax以post的方式传值,接收的值与传的不一样

利用ajax以post的方式传值,接收的值与传的不一样

Jun 06, 2016 pm 08:38 PM
ajax php post

在显示页面用ajax以post的方式传值,值data为json格式。在控制器中接收,但是只是接收到了部分值。传的值data中有个url,url中含有特殊符号&,接收的值就是到这个符号,后面的没有了。如截图,源代码中的data是我想要的值。利用ajax以post的方式传值,接收的值与传的不一样

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在显示页面用ajax以post的方式传值,值data为json格式。在控制器中接收,但是只是接收到了部分值。传的值data中有个url,url中含有特殊符号&,接收的值就是到这个符号,后面的没有了。如截图,源代码中的data是我想要的值。利用ajax以post的方式传值,接收的值与传的不一样

取决于你传递给data的类型:

  • 如果是对象,jquery会在发送ajax请求时自动添加encodeURIComponent,&符号被编码成%26,不会被formData自动截断
  • 如果data是一个字符串,jquery会自动将字符串放在formData里原封不动发送给服务器;由于formData是通过&截断的,所以字符串在没经过encodeURIComponent的&处自动被截断了

提交之前,URL Encode一下你传的这个url

data对应的值使用JSON.stringify({})封装一下即可
例如:

<code>$(".a_post").on("click",function(event){
    event.preventDefault();//使a自带的方法失效
    $.ajax({
           type: "POST",
           url: "url地址",
           contentType:"application/json",
           data: JSON.stringify({param1:"param1",param2:"param2",url:"http://xxxx?a=a&b=b&c=c",param3:"param3"}),//参数列表
           dataType:"json",
           success: function(result){
              //请求正确之后的操作
           },
           error: function(result){
              //请求失败之后的操作
           }
    });
});
</code>
Copy after login

查看请求参数如下:
利用ajax以post的方式传值,接收的值与传的不一样

用这个JSON.stringify({})封装了,就是ajax传的时候出问题了,用data:'data='+data,这样传值就有问题,但是用data:{data:data},就可以,是为什么呢利用ajax以post的方式传值,接收的值与传的不一样

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