如何通过ajax获取数据，将php变量传递给外部php文件？

问题描述：

I have 2 files, index.php and ajax.php, I am retrieving data from ajax.php to display in index.php via ajax.. I would like to pass the username to ajax.php to use in my sql statement.. How can I achieve this?

index.php -

$profile_user =$_GET['user'];

    $(document).ready(function(){
        $('.loader').hide();
        var load = 0;
        var nbr = "<?php echo $nbr;?>";
        $(window).scroll (function(){
            if($(window).scrollTop() == $(document).height() - $(window).height()){
                $('.loader').show();
                load++;
                if(load * 5 > nbr){
                    $('.messages').text("No more to see..");
                    $('.loader').hide();
                }else{

                $.post("ajax.php", {load:load},function(data){
                    $(".images").append(data);
                    $('.loader').hide();
                });
                }
            }
        });
    });

ajax.php

$profile_user =$_GET['user'];
$sql = "SELECT * FROM images WHERE  user = '$profile_user'  ORDER BY ID DESC ";

Can I pass the variable in the ajax call?

我有2个文件，index.php和ajax.php，我正在从ajax.php中检索数据以显示在 index.php通过ajax ..我想将用户名传递给ajax.php以在我的sql语句中使用..我怎样才能实现这个目的？ p>

index.php - p >

  $ profile_user = $ _ GET ['user']; 
 
 $（document）.ready（function（）{
 $（'。loader'）。hide（）  ; 
 var load = 0; 
 var nbr =“＆lt;？php echo $ nbr;？＆gt;”; 
 $（window）.scroll（function（）{
 if（$（window）.scrollTop  （）== $（document）.height（） -  $（window）.height（））{
 $（'。loader'）。show（）; 
 load ++; 
 if（load * 5＆gt;  nbr）{
 $（'。messages'）。text（“No more to see ..”）; 
 $（'。loader'）。hide（）; 
} else {
 
 $。  post（“ajax.php”，{load：load}，function（data）{
 $（“。images”）。append（data）; 
 $（'。loader'）。hide（）; 
  }）; 
  } 
} 
}）; 
}）; 
  code>  pre> 
 
  ajax.php  p> 
 
 
  $ profile_user  = $ _ GET ['user']; 
 $ sql =“SELECT * FROM images WHERE user ='$ profile_user'ORDER BY ID DESC”; 
  code>  pre> 
 
  Can 我在ajax调用中传递变量？ p> 
  div>

答

Add the PHP variable in the AJAX call

$.post("ajax.php", {
  load: load,
  user: '<?php echo $profile_user; ?>'
},function(data){

And change $_GET by $_POST in your ajax.php file, like this

$profile_user = $_POST['user'];

答

You need to POST that data:

            $.post('ajax.php', {load:load, user: '"<?php echo $_GET['user'];?>"'},function(data){
                $('.images').append(data);
                $('.loader').hide();
            });

答

You can change

$.post("ajax.php", {load:load},function(data){
                    $(".images").append(data);
                    $('.loader').hide();
                });

$.post("ajax.php", {load:load, user:user},function(data){
                        $(".images").append(data);
                        $('.loader').hide();
                    });

And in ajax.php after getting mysql result just encode the array in json format and echo it.

$data = mysql_query($sql);
echo json_encode($data);exit;

如何通过ajax获取数据，将php变量传递给外部php文件？

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