当$ v和$ plan大时，为什么INSERT不起作用？

Question

我的应用程序的想法是，我应该获得一个字符串，该字符串带有指向对它们进行排序的图像的链接。我的问题是，当我使用INSERT时，它会给我$ result [“ message”] =“ error”。我还注意到，如果列表很小，那么INSERT可以工作，这很奇怪。

<?php
if ($_SERVER['REQUEST_METHOD'] =='POST'){

$ids = $_POST['ids'];
$time = $_POST['time'];
$itemS = $_POST['item'];
$typeS = $_POST['type'];
require_once 'connect.php';


$v = explode(',', $itemS);
$plan =  explode(',', $typeS);

$ids_post = rand();

foreach ($plan as $key => $item) {
  if (trim($item) == 'i') {
      $rand = rand();
      $path = "posts_image/$rand.jpeg";
      $finalPath = "http://192.168.100.9/login/".$path;
      if (file_put_contents($path, base64_decode($v[$key]))) {
        $v[$key]  = $finalPath;
      } else {
        $result["message"] = "error";
        echo json_encode($result);
        mysqli_close($conn);
      }
    }

     if (trim($item) == 'ci') {
      $rand = rand();
      $path = "posts_image/$rand.jpeg";
      $finalPath = "http://192.168.100.9/login/".$path;
      if (file_put_contents($path, base64_decode($v[$key]))) {
        $v[$key]  = $finalPath;
      } else {
        $result["message"] = "error";
        echo json_encode($result);
        mysqli_close($conn);
      }
    }
  }

      $item =  implode(',', $v);
      $type = implode(',', $plan);

      if (mysqli_query($conn, "INSERT INTO `new_post` (ids_post, ids, item, type, time_post ) VALUES ('$ids_post', '$ids', '$item', '$type', '$time')")) {
          $result["message"] = "success";
          echo json_encode($result);
          mysqli_close($conn);
      } else {
          $result["message"] = "error";
          echo json_encode($result);
          mysqli_close($conn);
      }
} else {
  $result["message"] = "error";
  echo json_encode($result);
  mysqli_close($conn);
}