php - 解析多个json对象并将其存储在数据库中

Question

我需要从 JSON 文件中解析这多个 JSON 对象。 JSON 文件包含数千个 JSON 对象，现在我需要解析所有这些 JSON 对象，所以请任何人尝试解决这个问题，然后我再次需要将所有 JSON 数据存储到 MySQL。

{
  {
    "name": "Michael Bruce",
    "gender": "Male",
    "designation": "System Architect"
  },
  {
    "name": "Jennifer Winters",
    "gender": "Female",
    "designation": "Senior Programmer"
  },
  {
    "name": "Donna Fox",
    "gender": "Female",
    "designation": "Office Manager"
  },
  {
    "name": "Howard Hatfield",
    "gender": "Male",
    "designation": "Customer Support"
  }
}

{
  {
    "name": "aMichael Bruce",
    "gender": "Male",
    "designation": "System Architect"
  },
  {
    "name": "Jennifer Winters",
    "gender": "Female",
    "designation": "Senior Programmer"
  },
  {
    "name": "Donna Fox",
    "gender": "Female",
    "designation": "Office Manager"
  },
  {
    "name": "Howard Hatfield",
    "gender": "Male",
    "designation": "Customer Support"
  }
}

{
  {
    "name": "Michael Bruce",
    "gender": "Male",
    "designation": "System Architect"
  },
  {
    "name": "Jennifer Winters",
    "gender": "Female",
    "designation": "Senior Programmer"
  },
  {
    "name": "Donna Fox",
    "gender": "Female",
    "designation": "Office Manager"
  },
  {
    "name": "Howard Hatfield",
    "gender": "Male",
    "designation": "Customer Support"
  }
}

以及我的代码。

<?php

$host     = "localhost";
$username = "root";
$password = "";
$dbname   = "employee";
$con = mysqli_connect($host, $username, $password, $dbname) or die('Error in Connecting: ' . mysqli_error($con));

$st = mysqli_prepare($con, 'INSERT INTO emp(name, gender, designation) VALUES (?, ?, ?)');

mysqli_stmt_bind_param($st, 'sss', $name, $gender, $designation);

$filename = 'file3.json';
$json     = file_get_contents($filename);
$data = json_decode($json, true);

foreach ($data as $key => $value) {
    if (!is_array($value)) {
        echo $key . '=>' . $value . '<br/>';
    } else {
        foreach ($value as $key => $val) {

            $name        = $key['name'];
            $gender      = $key['gender'];
            $designation = $key['designation'];
            echo $name;
            mysqli_stmt_execute($st);
        }
    }
}

mysqli_close($con);
?>

上面的代码仅适用于一个 JSON 对象，假设我的文件中有多个 JSON 对象，然后它会抛出类似的错误。

Warning: Invalid argument supplied for foreach() in C:\xampp\htdocs\jumadi\b\c\index2.php on line 25

Answer 1

首先json是错误的。外括号必须是方形的，因为这应该是数组。

其次，检索代码从 key 读取而不是从 val 读取。应该是:

 $name = $val['name'];
  ... etc