此代码工作,但我需要使用json_encode如何使用json_encode?
<?php
require_once('person_class.php');
$person = new Person_class();
$first_name = addslashes ($_POST['first_name']);
$last_name = addslashes ($_POST['last_name']);
$birthday = addslashes ($_POST['birthday']);
$gender = addslashes ($_POST['gender']);
$person_id = $person->addPerson($first_name, $last_name, $birthday, $gender);
echo "
{ \"status\" : \"1\",
\"error\" : \"0\",
\"person_id\" : \"$person_id\",
\"first_name\" : \"$first_name\",
\"last_name\" : \"$last_name\",
\"birthday\" : \"$birthday\",
\"gender\" : \"$gender\"
}";
?>
我想改变这个部分,并使用json_encode
回声 “ {\” 状态\”:\ “1 \”, “error \”:\“0 \”, \“person_id \”:\“$ person_id \”, \“first_name \”:\“$ first_name \”, \“last_name \”:\“ $ last_name \“, \”birthday \“:\”$ birthday \“, \”gender \“:\”$ gender \ “ }”;
你的问题不清楚,你可以发布你想要的结果吗? –
为什么你先编码$ person_id并赋值给'$ json'变量,然后你重新声明'$ json' var? –