无法使用PHP从JSON文件向MYSQL数据库添加数据

I'm new with PHP. I'v been struggling with this task for hours now. Earlier I used json_encode to get data from MYSQL to a JSON file. Now i try to revese and add the same data from JSON file to a new MYSQL database. I have a problem with converting the array to string before passing it to MYSQL database tho. The database works and I was able to add "players" there by inserting manual values instead of the $values from array. My code looks like this:

我是PHP的新手。我现在几个小时都在努力完成这项任务。之前我使用json_encode将数据从MYSQL获取到JSON文件。现在我尝试将JSON文件中的相同数据添加到新的MYSQL数据库中。我有一个问题，在将数组传递给MYSQL数据库之前将数组转换为字符串。数据库工作，我能够通过插入手动值而不是数组中的$值来添加“播放器”。我的代码如下所示：

<?php
//open connection to mysql db
$con = mysqli_connect("localhost","root","","scoreboard2") or die("Error " . mysqli_error($con));

$scorefile = file_get_contents('scores.json');
$Score = json_decode($scorefile, true);

echo '<pre>' . print_r($Score, true) . '</pre>';

foreach ($Score as $field => $value) {
    // Use $field and $value here

    print_r($field . '=>' . $value . '<br/>', true);

    //mysqli_query($con, "INSERT INTO scores (name, score, time) VALUES ($value->name, $value->score, $value->time)");
}

//mysqli_close($con);
?>

the JSON file looks like this:

JSON文件如下所示：

[
 {"id":"22",
  "name":"Jack",
  "score":"2142",
  "time":"196:13",
  "ts":"2016-02-23 15:36:23",
  "date":"2016-02-23"},

 {"id":"23",
  "name":"Bob",
  "score":"7026",
  "time":"35:54",
  "ts":"2016-02-23 15:40:33"}
]

etc.. and the "error" is this:

等等。“错误”是这样的：

Notice: Array to string conversion in F:\XAMPP\htdocs\JSON_MySQL\decode.php on line 13

注意：第13行的F：\ XAMPP \ htdocs \ JSON_MySQL \ decode.php中的数组到字符串转换

2 个解决方案

#1

Your issue is that you are converting the data in your .json file to an array by using parameter 2 of json_decode() as true.

您的问题是您使用json_decode（）的参数2将.json文件中的数据转换为数组。

This is converting your objects to arrays and therefore the syntax you use in this line is wrong

这是将您的对象转换为数组，因此您在此行中使用的语法是错误的

mysqli_query($con, "INSERT INTO scores 
                           (name, score, time) 
                    VALUES ($value->name, $value->score, $value->time)");

because you are using object notation.

因为你正在使用对象表示法。

So change this line from

所以改变这一行

$Score = json_decode($scorefile, true);

To

至

$Score = json_decode($scorefile);

SO

所以

?php
$con = mysqli_connect("localhost","root","","scoreboard2") or die("Error " . mysqli_error($con));

$scorefile = file_get_contents('scores.json');
$Score = json_decode($scorefile);

foreach ($Score as $object) {

    mysqli_query($con, "INSERT INTO scores 
                               (name, score, time) 
                        VALUES ('{$object->name}', '{$object->score}', '{$object->time}')");
}

mysqli_close($con);
?>

Also note I quoted the values with single quotes and also wrapped the object properties in {} which is required when using object or array notation inside a double quoted literal.

另请注意，我使用单引号引用值，并在{}中包装对象属性，这在双引号文字中使用对象或数组表示法时是必需的。

#2

The error

错误

Notice: Array to string conversion in F:\XAMPP\htdocs\JSON_MySQL\decode.php on line 13

注意：第13行的F：\ XAMPP \ htdocs \ JSON_MySQL \ decode.php中的数组到字符串转换

is produced by the line

是由线产生的

print_r($field . '=>' . $value . '<br/>', true);

print_r（$ field。'=>'。$ value。'
'，true）;

(which actually is the 13th line)

（实际上是第13行）

where you try to convert $value which is an array (your second score object) to a string in order to concatenate it with the rest of the string

在哪里尝试将$ value（一个数组（您的第二个得分对象））转换为字符串，以便将其与字符串的其余部分连接起来

note that if you replace the error-producing line by

请注意，如果您替换错误生成行

echo '<pre>' .$field . '=>' . print_r($value, true) . '</pre>'.'<br/>';

you get the

你得到了

0=>Array
(
    [id] => 22
    [name] => Jack
    [score] => 2142
    [time] => 196:13
    [ts] => 2016-02-23 15:36:23
    [date] => 2016-02-23
)

1=>Array
(
    [id] => 23
    [name] => Bob
    [score] => 7026
    [time] => 35:54
    [ts] => 2016-02-23 15:40:33
)

that you might originally expect

你可能原本期望的