Javascript PHP MySQL查询在实际提交的行之前插入3行（来自w2ui表）

我收到一个表的请求，要求从表单向数据库中插入数据。 我能够成功提交数据，但每次我提交一份表格时，它都会提交3个其他条目

我尝试了

stripcslashes

并转义字符串，但这不是问题所在

以下是未修剪请求的外观：

request=%7B%22cmd%22%3A%22save%22%2C%22recid%22%3A0%2C%22name%22%3A%22foo%22%2C%22record%22%3A%7B%22imei%22%3A%22doe%22%2C%22mobile_no%22%3A%22Doe%22%2C%22iccid%22%3A%22jdoe%40email.com%22%2C%22driver%22%3A%22jdoe%22%2C%22company%22%3A%22jdoe%40email.com%22%2C%22type%22%3A%22jdoe%40email.com%22%2C%22channels%22%3A%22jdoe%40email.com%22%2C%22cameras%22%3A%22jdoe%40email.com%22%7D%7D

以下是修剪请求的外观：

request=%7B%22cmd%22%3A%22save%22%2C%22recid%22%3A0%2C%22name%22%3A%22foo%22%2C%22record%22%3A%7B%22imei%22%3A%22doe%22%2C%22mobile_no%22%3A%22Doe%22%2C%22iccid%22%3A%22jdoe%40email.com%22%2C%22driver%22%3A%22jdoe%22%2C%22company%22%3A%22jdoe%40email.com%22%2C%22type%22%3A%22jdoe%40email.com%22%2C%22channels%22%3A%22jdoe%40email.com%22%2C%22cameras%22%3A%22jdoe%40email.com%22%7D%7D

我将字段填入

0

进行测试

{"cmd":"save","recid":0,"name":"foo","record":{"imei":"0","mobile_no":"0","iccid":"0","driver":"0","company":"0","type":"0","channels":"0","cameras":"0"}}

为了得到Json格式，必须对请求进行修剪和处理

以下是我提交请求的PHP代码：

<?php
//get request a put into variable
$input = file_get_contents('php://input');
$decoded = urldecode($input);
//trim "request=" off
$Trimmed = trim($decoded, 'request=');

//con vars
$servername = "servername";
$username = "username";
$password = "password";
$dbname = "dbname";
//time vars
$date_added1 = date("Y-m-d");
$time_added1 = date("h:i:sa");

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);

$array = json_decode($Trimmed, true);

foreach($array as $row)
{
    $sql = "INSERT into `adddevices` (imei, mobile_no, iccid, driver, company, type, channels, date_added, time_added, cameras) 
VALUES ('".$row["imei"]."', '".$row["mobile_no"]."', '".$row["iccid"]."', '".$row["driver"]."', '".$row["company"]."', '".$row["type"]."', '".$row["channels"]."', '$date_added1', '$time_added1', '".$row["cameras"]."')";

mysqli_query($conn, $sql);

}
echo "Data inserted";
mysqli_close($conn);

?>

---

您正在
$array
上运行
foreach（）。然后你就可以放下餐车去做

$row = $array['record'];

谢谢你的回答，最后我一起修改了代码，并使用了
$jsonArray=json_decode（$jsonString，true）['record']
要获取记录而不是foreach循环，我的代码不会工作，因为我误解了。你的新代码应该可以工作，但我已经更新了我的答案，以防有人从搜索引擎过来，我可以用原始代码来解决这个问题