php非法字符串偏移量中遇到错误
我正在编写一个php代码,从伪json api服务器获取数据。但是我面临这个错误php非法字符串偏移量中遇到错误,php,arrays,json,Php,Arrays,Json,我正在编写一个php代码,从伪json api服务器获取数据。但是我面临这个错误 Warning: Illegal string offset 'employee_name' in C:\xampp\htdocs\jsonapi.php on line 19 name: s Warning: Illegal string offset 'employee_age' in C:\xampp\htdocs\jsonapi.php on line 21 name: s Notice: Undef
Warning: Illegal string offset 'employee_name' in C:\xampp\htdocs\jsonapi.php on line 19
name: s
Warning: Illegal string offset 'employee_age' in C:\xampp\htdocs\jsonapi.php on line 21
name: s
Notice: Undefined index: employee_name in C:\xampp\htdocs\jsonapi.php on line 19
name:
Notice: Undefined index: employee_age in C:\xampp\htdocs\jsonapi.php on line 21
name:
这是我的密码
<?php
$api_url = 'http://dummy.restapiexample.com/api/v1/employees';
// Read JSON file
$json_data = file_get_contents($api_url);
// Decode JSON data into PHP array
$user_data = json_decode($json_data,true);
// Cut long data into small & select only first 10 records
$user_data = array_slice($user_data,0,9);
// Print data if need to debug
//print_r($user_data);
// Traverse array and display user data
foreach ($user_data as $user) {
echo "name: ".$user['employee_name'];
echo "<br />";
echo "name: ".$user['employee_age'];
echo "<br /> <br />";
}
?>
如何删除此类错误?正如您在中所看到的,实际数据包含在数据
字段中
因此,请替换:
$user_data = json_decode($json_data, true);
与:
这意味着
$user\u data
包含字符串,而不是对象。我该怎么办?不客气。请考虑通过单击✔ 左边的符号。
$result = json_decode($json_data, true);
$user_data = $result['data'];