php - 使用 fgetcsv() 发出解析数据 - 期望参数为资源

Question

在 php 中使用 fgetcsv() 解析来自 URL 的数据时出现此错误。

这是我的错误:PHP 警告:fgetcsv() 期望参数 1 为资源， bool 值在第 20 行的/testing.php 中给出

注意:我已经厌倦了一个没有运气的可能解决方案 PHP return error when using fgetcsv

数据集:我试图用 fgetcsv() 解析的内容

{"pos":[{"pos":"1","person":"John","place":"8","score":"109","round":"1"}, {"pos":"2","person":"Mary","score":"80"},{"pos":"3","person":"Luke Guthrie","place":"7*","score":"7","round":"1"}, {"pos":"4","person":"Charles","score":"3"},{"pos":"5","person":"Sean","place":"8","score":"4","round":"1"}, {"pos":"6","person":"Rob","score":"60"},   {"pos":"7","person":"Dan","score":"76"},{…

这是我的 php 文件:

<?php

$servername = "localhost";
$username = "root";
$password = "xxxx";
$dbname = "player";

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

  $f_pointer=fopen("URL/api/v2/projects/xxx/last_ready_run/data","r"); 



  while(! feof($f_pointer))

  {
      $ar=fgetcsv($f_pointer);


 $sql="INSERT INTO scores(pos,person,score,place,score,round)values('$ar[0]','$ar[1]','$ar[2]','$ar[3]','$ar[4]','$ar[5]')";
 $conn->query($sql); 
 echo $sql;
 echo "<br>";


  }
 ?>

Answer 1

再试一些错误处理。

$f_pointer=fopen("URL/api/v2/projects/xxx/last_ready_run/data","r"); 
if ( !$f_pointer ) {
  trigger_error('fopen failed', E_USER_ERROR);
}

无论如何，fgetcsv() 不是解析 json 的正确函数。 json_decode()是。