【发布时间】:2017-10-02 13:19:35
【问题描述】:
对于一个学校项目,我想遍历 JSON 响应并将数据插入 MySQL 数据库。
JSON 响应如下所示:
{"tableID":100965,"data":[{"updated_at":1506994152000,"prices":{"unstable_reason":"LOW_SALES","unstable":true,"sold":{"avg_daily_volume":null,"last_90d":67,"last_30d":52,"last_7d":15,"last_24h":1},"max":287.5,"avg":93.52,"min":51.01,"latest":87},"image":"IMGURL","db_name":"Test-Artikel","dbID":"123456789"},{"updated_at":1506994152000,"prices":{"unstable_reason":"LOW_SALES","unstable":true,"sold":{"avg_daily_volume":null,"last_90d":67,"last_30d":52,"last_7d":15,"last_24h":1},"max":287.5,"avg":93.52,"min":51.01,"latest":87},"image":"IMGURL","db_name":"Test-Artikel","dbID":"123456789"}],"__v":0,"createdAt":"2017-03-27T09:16:48.395Z"}
每个数组都应该是 mysql 数据库中自己的一行。
目前我可以使用以下代码从 json 文件中获取一个信息:
<?php
$url = "linktojson";
//connect to database
//$pdo = new PDO('mysql:host=localhost;dbname=...', '...', '...');
//read the json file contents
$jsondata = file_get_contents($url);
//convert json object to php associative array
$data = json_decode($jsondata, true);
echo('<pre>');
foreach ($data['data'] as $api_data) {
echo $api_data['name'] . '<br/>';
}
?>
编辑: 当前代码:
$pdo = new PDO('mysql:host=localhost;dbname=XXX', 'XXX', 'XXX');
$yourJsonArray ="test.json";
$dataArray = json_decode(json_encode($yourJsonArray),true);
foreach($dataArray as $key => $value){
$image = $key['image'];
$statement = $pdo->prepare('INSERT INTO api_data (image) VALUES (?)');
$statement->execute(array($image));
}
错误消息:警告:在第 29 行的 test.php 中为 foreach() 提供的参数无效
第 29 行 = foreach(顶部有一些旧的 cmets)
编辑:在 foreach 之前添加 var_dump($yourJsonArray);:
string(638) "{"tableID":100965,"data":[{"updated_at":1506994152000,"prices":{"unstable_reason":"LOW_SALES","unstable":true,"sold":{"avg_daily_volume":null,"last_90d":67,"last_30d":52,"last_7d":15,"last_24h":1},"max":287.5,"avg":93.52,"min":51.01,"latest":87},"image":"IMGURL","db_name":"Test-Artikel","dbID":"123456789"},{"updated_at":1506994152000,"prices":{"unstable_reason":"LOW_SALES","unstable":true,"sold":{"avg_daily_volume":null,"last_90d":67,"last_30d":52,"last_7d":15,"last_24h":1},"max":287.5,"avg":93.52,"min":51.01,"latest":87},"image":"IMGURL","db_name":"Test-Artikel","dbID":"123456789"}],"__v":0,"createdAt":"2017-03-27T09:16:48.395Z"}"
Warning: Invalid argument supplied for foreach() in test.php on line 30
编辑:将 $dataArray 更改为 $dataArray = json_decode($yourJsonArray,true);
错误 (4x): 警告:第 33 行 test.php 中的非法字符串偏移“图像”
第 33 行 = $image = $key['image'];
编辑:我转储了 $key Var 并得到了这个:
string(7) "tableID" string(4) "data" string(3) "__v" string(9) "createdAt"
如何获取“数据”的元素?
【问题讨论】:
-
我知道如何循环遍历文件和每个数组中的一个元素(例如“名称”)。回声('
'); foreach ($data['data'] as $name) { echo $name['name'] 。 '
';我找不到合适的方法来获取我需要的所有元素。 -
你有'json响应'看起来像这样`'但那不是json。把真实的东西放在人们可以帮助你的地方,而不是你认为你得到的东西,或者你认为你在'linktojson'上生产的东西。
-
好吧。我明白你的意思 - 将 json 文件的前两个元素添加到我的问题中。
-
ahhh ...他的眼睛已经看到了(并且jsonlint认为这个json是有效的)!因此,
$api_data数组似乎没有name属性。你是说db_name吗? -
哈哈.. 我的问题中的 php 代码“已过时”-我现在使用 db_name。
标签: php mysql json loops foreach