PHP JSON解析解决办法
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2022-05-14 14:54:55
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PHP JSON解析
up_user_info={"name":"liux","sex":"1","phone":"13712800254","email":"[email protected]","town":"town_1","age":"18","heigh":"180","experience":"工作经验"}
$up_user_info = $_POST['up_user_info'];
if($up_user_info == null){
$error = 1;
}
$upuser_decode = json_decode($up_user_info,true);
$name = $upuser_decode['name'];
echo $name;
输出为空~~为什么...想来想去都不知道哪里有错!
------解决方案--------------------
你是说 $_POST['up_user_info'] = '{"name":"liux","sex":"1","phone":"13712800254","email":"[email protected]","town":"town_1","age":"18","heigh":"180","experience":"工作经验"}';
print_r(json_decode($_POST['up_user_info'])); 为空？
那说明你是在 gbk 环境下
print_r(json_decode(iconv('gbk', 'utf-8', $_POST['up_user_info'])));
就可以了
------解决方案--------------------
你可以這樣測試
1.$up_user_info是post過來的
2.使用我上面的程序
分別測試是否可以輸出。
如果1不行 2可以，就是POST過來的數據有問題，請檢查這裡的數據。
up_user_info={"name":"liux","sex":"1","phone":"13712800254","email":"[email protected]","town":"town_1","age":"18","heigh":"180","experience":"工作经验"}
$up_user_info = $_POST['up_user_info'];
if($up_user_info == null){
$error = 1;
}
$upuser_decode = json_decode($up_user_info,true);
$name = $upuser_decode['name'];
echo $name;
输出为空~~为什么...想来想去都不知道哪里有错!
------解决方案--------------------
你是说 $_POST['up_user_info'] = '{"name":"liux","sex":"1","phone":"13712800254","email":"[email protected]","town":"town_1","age":"18","heigh":"180","experience":"工作经验"}';
print_r(json_decode($_POST['up_user_info'])); 为空？
那说明你是在 gbk 环境下
print_r(json_decode(iconv('gbk', 'utf-8', $_POST['up_user_info'])));
就可以了
stdClass Object
(
[name] => liux
[sex] => 1
[phone] => 13712800254
[email] => [email protected]
[town] => town_1
[age] => 18
[heigh] => 180
[experience] => 工作经验
)
------解决方案--------------------
你可以這樣測試
1.$up_user_info是post過來的
2.使用我上面的程序
分別測試是否可以輸出。
如果1不行 2可以，就是POST過來的數據有問題，請檢查這裡的數據。
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