doudi4137 2017-10-24 09:54
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如何通过cURL获取数据并等待结果?

I have a list of IDs. For test this is about 20 items.

Now I want to make a loop and get extra data from another server. On this server I prepared a script - when you send it an item ID it will send back more details in JSON format.

while($row = mysql_fetch_array($result)){   
    $ch = curl_init();
    curl_setopt($ch, CURLOPT_URL, "http://test.mysite.abc/call/itemdetail/id/".$row['id_item']);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    $ret = curl_exec ($ch);
    curl_close ($ch);

    $result = json_decode($ret, true);

    print_r($result);
    vecho "<br>";
}

When I go on mysite.abc I see result when I give an item id. But when I run script sometimes I have 1 result or 3 results and the message:

Warning: mysql_fetch_array() expects parameter 1 to be resource, array given in C:\wamp64\www\testfile.php on line 65

How should I do this?

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1条回答 默认 最新

  • dsf45346 2017-10-24 13:23
    关注

    Check mysql_query returned a valid resource type.

    $result = mysql_query('SELECT column1, column2 FROM table WHERE 1=1'); // Change the SQL query with your
    
    if ($result) {
        while($row = mysql_fetch_array($result)){   
            $ch = curl_init();
            curl_setopt($ch, CURLOPT_URL, "http://test.mysite.abc/call/itemdetail/id/".$row['id_item']);
            curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
            $ret = curl_exec ($ch);
            curl_close ($ch);
    
            $result = json_decode($ret, true);
    
            print_r($result);
            echo "<br>";
        }
    }
    

    Best Practices:, Avoid mysql_* extension. This extension was deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0. Instead, use the MySQLi or PDO_MySQL extension.

    Consider using multi-curl for your case. Look this answer for more details.

    Cheerse!!

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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