鍛鍊與PHP返回數組jQuery的

Question

我這個PHP代碼工作爲$就調用，它是下面這段代碼

$family = mysql_real_escape_string($_REQUEST['send_txt'], $link); 
$query = "SELECT imgurl FROM images WHERE family='$family'"; 

//Query database 
$result = mysql_query($query, $link); 

//Output result, send back to ajax as var 'response' 

$imgurl=array(); 
//$i=0; 
if(mysql_num_rows($result) > 0){ 
    //Fetch rows 
    while($row = mysql_fetch_array($result)){ 

     $imgurl[]=$row['imgurl']; 

     } 
} 
echo $imgurl; 

jQuery代碼

$(document).ready(function() { 
$('ul.sub_menu a').click(function() { 
    var txt = $(this).text(); 
     $.ajax({ 
     type: "POST", 
     url: "thegamer.php", 
     data:{send_txt: txt}, 
     success: function(data){ 
     $('#main-content').html(data); 
     } 
     }); 
    }); 
}); 

其輸出只是在陣列寫＃main-content div如何處理基本爲圖像路徑的數組

Answer 1

爲什麼你創建的MySQL結果陣列？你的代碼可以是這樣的簡單：

<?php 
$family = mysql_real_escape_string($_REQUEST['send_txt'], $link); 
$query = "SELECT imgurl FROM images WHERE family='$family'"; 

//Query database 
$result = mysql_query($query, $link); 

//Output result, send back to ajax as var 'response' 

if(mysql_num_rows($result) > 0) 
{ 
    //Fetch rows 
    while($row = mysql_fetch_array($result)) 
    { 
     echo $row['imgurl']; 
    } 
} 
?> 

Answer 2

直接從瀏覽器嘗試頁面。使用JSON可以幫助在這裏：

echo json_encode($imgurl); 

和使用getJSON不是純ajax：

$.getJSON('thegamer.php', {send_text:text}, function(data) { … });