json_encode 问题？

juliets 2010-03-05 01:24:50

$arr = array(
'uname' => '张三',
'csdn' => 'https://forum.csdn.net/',

);

$code = json_encode($arr);
$json = preg_replace("#\\\u([0-9a-f]+)#ie", "iconv('UCS-2', 'UTF-8', pack('H4', '\\1'))", $code);

echo $json;

输出以下{"uname":"张三","csdn":"http:\/\/forum.csdn.net\/"}

/前面自动加上了个\怎么去掉呢
按原地址输出！

...全文

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lsy51213 2010-11-25

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[Quote=引用 11 楼 zyu_2005 的回复:]
引用 3 楼 kyzy_yy_pm 的回复:

echo stripcslashes($path); //php自带函数也可以

你用这个就行
[/Quote]
这个行!

casahama 2010-05-26

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[Quote=引用 10 楼 nq19860409 的回复:]
该怎么解决啊帮忙
<?php

$servname="localhost";
$sqlservname="root";
$sqlservpws="123456";
$sqlname="lock1";
$db=mysql_connect($servname,$sqlservname,$sqlservpws) or die("数据库连接失败");
mysql_select_db……
[/Quote]

看你頁面編碼和數據庫編碼是否一致.

zyu_2005 2010-04-23

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[Quote=引用 3 楼 kyzy_yy_pm 的回复:]

echo stripcslashes($path); //php自带函数也可以
[/Quote]
你用这个就行

nq19860409 2010-03-31

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该怎么解决啊帮忙
<?php

$servname="localhost";
$sqlservname="root";
$sqlservpws="123456";
$sqlname="lock1";
$db=mysql_connect($servname,$sqlservname,$sqlservpws) or die("数据库连接失败");
mysql_select_db($sqlname,$db);
$sql = "select * from t_operater";
$result =mysql_query($sql);
$rows = mysql_num_rows($result);
while($obj = mysql_fetch_object($result))
{
$arr[] = $obj;

}

echo '({"total":"'.$rows.'","results":'.json_encode($arr).'})';
?>

运行结果是
({"total":"204","results":[{"operater_id":"-9999","org_id":null,"phone":null,"name":"??IE??","pwd":"-9999","state":"A"},{"operater_id":"1010","org_id":"2","phone":"15316707210","name":"??10","pwd":"10","state":"A"},{"operater_id":"1111","org_id":"2","phone":"15316707205","name":"???","pwd":"111111","state":"X"},{"operater_id":"9999","org_id":"2","phone":"15316707209","name":"??9","pwd":"111111","state":"A}）

name是乱码