doswy02440 2015-10-27 19:59
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Ajax - jQuery和Ajax错误/无效

I am building a webpage that contains a div which holds some data retrieved from a MySql database through PHP. This div shows some products and on its left there is a simple nav bar with different categories. What I am attempting to do is that when the user clicks a category of this nav bar, the content of the div will change, showing the products of that category all of which are stored in the db. So I tried using Ajax (my first time btw), and I can't seem to make it work. My project structure is something like this:

Parent directory

  • php > index.php
  • css > index.css
  • js > index.js
  • img > images here
  • ajax> products-ajax.js / products-ajax.php

The index.php file is linked to both index.js AND products-ajax.js However, I have already tried including the Ajax line of code in both index.js and index.php but I can't make it recieve data back from products-ajax.php Any help is appreciated.

And here's what my test code looks like:

/* THIS IS THE products-ajax.js */

    $('.products-list li').click(function(){

               {p: "Product name"},
               success: function(data){alert(data)}

/* THIS IS THE products-ajax.php */

$p = $_POST['p'];

echo $p;


I realized my broswer's debugger says there's a missing parentheses:

$('.products-list li').click(function(){
 //The debugger says HERE v should go a parentheses
$.post("products-ajax.php", {p: "Product name"}, success: function(data{alert(data)});

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1条回答 默认 最新

  • dousong1992 2015-10-27 20:21

    Using $.ajax() instead of $.post(), you could add an error handler which would give you precious information (maybe you can with $.post(), but I'm not familiar with this function, which afaik is just a shortcut to $.ajax()). It could be something like :

    var request = $.ajax({
        url: "products-ajax.php",
        method: "POST",
        data: { p: "Product name" }
    request.done(function( data ) {
        alert( data );
    }); jqXHR, textStatus ) {
        alert( "Request failed: " + textStatus );

    Edit: and prefer console.log() rather than alert() for debugging. Especially with asynchronous interactions, alert() sometimes leads to surprises...

    本回答被题主选为最佳回答 , 对您是否有帮助呢?



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