我想從我的數據庫獲取圖像路徑的列表,並在Jquery和Json triying的幫助下添加到我的網站。但我不t know why after encoding my string using
json_encode`在PHP它改變它的路徑,並顯示我像Jquery:JSON更改路徑的字符串
[{"0":"user\/photogallery\/images\/members\/2\/2_1.jpg","src":"user\/photogallery\/images\/members\/2\/2_1.jpg"},{"0":"user\/photogallery\/images\/members\/2\/2_2.jpg","src":"user\/photogallery\/images\/members\/2\/2_2.jpg"}]
我只需要user/photogallery/images/members/2/2_2.jpg
部分來創建新的<img src ="user/photogallery/images/members/2/2_2.jpg " />
組件。
這裏我的PHP代碼和腳本
$member_id = $GET['member_id'];
$files = find_all_photos($member_id);
$encoded = json_encode($files);
echo $encoded;
unset($encoded);
function find_all_photos($id)
{
db_connect();
$query = sprintf("SELECT src FROM photo_album_list WHERE user_id = '%s'",
mysql_real_escape_string($id));
$result = mysql_query($query);
$result = db_result_to_array($result);
return $result;
}
function db_result_to_array($result)
{
$res_array = array();
for ($count=0; $row = mysql_fetch_array($result); $count++)
{
$res_array[$count] = $row;
}
return $res_array;
}
和文字
$.get('photostack.php', {member_id:2} , function(data) {
console.log(data);
var items_count = data.length;
for(var i = 0; i < items_count; ++i){
var item_source = data[i];
var cnt = 0;
$('<img />').load(function(){
var $image = $(this);
++cnt;
resizeCenterImage($image);
$ps_container.append($image);
var r = Math.floor(Math.random()*41)-20;
if(cnt < items_count){
$image.css({
'-moz-transform' :'rotate('+r+'deg)',
'-webkit-transform' :'rotate('+r+'deg)',
'transform' :'rotate('+r+'deg)'
});
}
if(cnt == items_count){
$loading.remove();
$ps_container.show();
$ps_close.show();
$ps_overlay.show();
}
}).attr('src',item_source);
}
},'json');
更多的PHP問題比jQuery的問題。我已經添加了PHP標籤,並且您可能會考慮刪除jQuery。 – 2011-02-27 10:28:18
你是否從服務器端獲得它?檢查你的控制檯上的響應 – 2011-02-27 10:28:31
可能的重複[JSON:爲什麼是正斜槓?](http://stackoverflow.com/questions/1580647/json-why-are-forward-slashes-escaped) – 2011-02-27 10:31:53