当我尝试使用 //query1
$sql = " DELETE FROM lkeuser WHERE userid = '$sew' AND songid = '$a' " ;
$conn->query($sql);
//query2
$sql = "UPDATE liking
SET count = count - 1 ";
if ($conn->query($sql) === TRUE) {
函数从网址http://example.com/img.jpg?aa=11&bb=22
获取图片时,网址编码为file_get_contents()
。
它显示错误http://example.com/img.jpg?aa=11&bb=22
如何避免此网址编码?
我的代码是
file_get_contents(http://example.com/img.jpg?aa=11&bb=22) : failed to open stream: Connection timed out in filename.php
答案 0 :(得分:0)
我不确定编码的来源,但您可以使用List<String>
对其进行解码:
html_entity_decode()