Home > Backend Development > PHP Tutorial > Ajax quickly solves the problem that the parameter is too long and cannot be submitted successfully. How to solve it

Ajax quickly solves the problem that the parameter is too long and cannot be submitted successfully. How to solve it

小云云
Release: 2023-03-19 12:16:01
Original
2656 people have browsed it

This article mainly brings you an ajax quick solution to the problem of parameters that are too long and cannot be submitted successfully. The editor thinks it is quite good, so I will share it with you now and give it as a reference for everyone. Let’s follow the editor to take a look, I hope it can help everyone.

I checked a lot of information and said that the parameters of the get method are limited, but the length of the parameters of the post method is unlimited. This is also the advantage of post over get.

Use the post method in ajax and use the conventional parameter format: param1=a1¶m2=a2. When the parameter length is too long, the submission is still unsuccessful. For example, we often write an ajax post request like this:

$.ajax({
  type: "post",  // post or get
  contentType:"application/json;charset=utf-8",
  data: “requestTag=”+tag+"&content="+content,  //请求参数
  url: "postMockJson",   //地址
  dataType: "text",
  error: function (err) {
   outLog("错误"+err);
  },
  success: onSaveSuccess
});
Copy after login

When used like this, we find that if parameter 2: content is too much, for example, I am passing a relatively large text content, and I am serving it in the background ( When I get it from servlet):

String content= request.getParameter("content");

The value of content here is null.

There is also a quick way to check whether the ajax request is successful. Use the F12 developer tool to debug. After executing the ajax code, you can see the initiated request in the network options page in the F12 tool. , the requested parameters seen at this time have error prompts.

Solution:

There is another way to write the parameter format of ajax: the request parameter in json format, I can write it like this:

var param = "{requestTag: \""+requestTag+"\",content:\""+content+"\"}";
(ps: pay attention to the correct json format)

At this time, if you use F12 for debugging, you can The data in the requested parameters are correct.

Then the question is, the content I get in the servlet is still null. Why is this? ? ?

Since the request parameter is a json data block, of course this request.getParameter("content") method cannot obtain the data because it will not parse the json data for us.

So where is the parameter data we passed?

Here’s the point: the data is all in the request object.

Then we use the most primitive method to obtain the transferred data through the data flow method, as follows:

request.setCharacterEncoding("UTF-8");
StringBuilder sb = new StringBuilder();
try(BufferedReader reader = request.getReader();) {
char[] buff = new char[1024];
int len;
 while((len = reader.read(buff)) != -1) {
  sb.append(buff,0, len);
  }
}catch (IOException e) {
   e.printStackTrace();
}
Copy after login

At this time, our json data is all in the sb object , then you only need to parse the json object:

JSONObject jobject = JSONObject.fromObject(sb.toString());
String requestTag = jobject.getString("requestTag");
String content = jobject.getString("content");
Copy after login

At this point, we can get the content.

Related recommendations:

The value of the Ajax submission parameter contains the html tag and cannot be submitted successfully. How to solve the problem

Ajax files and other Parameter upload function

JavaScript parameter transfer illustration tutorial

The above is the detailed content of Ajax quickly solves the problem that the parameter is too long and cannot be submitted successfully. How to solve it. For more information, please follow other related articles on the PHP Chinese website!

Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template