【问题标题】:MySQL returned an empty result setMySQL 返回一个空的结果集
【发布时间】:2015-01-06 14:36:14
【问题描述】:

我不明白为什么 mysql 表返回空结果集,代码似乎还可以。 来自 mysql 的响应:MySQL 返回了一个空的结果集(即零行)。 (查询耗时 0.0010 秒)

<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "myexcel";

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


if (isset($_POST['submit']))
{
        $file = $_FILES['file']['tmp_name'];
        $handle = fopen( $file, "r");
        while(($fileop = fgetcsv($handle, 3000, ",")) !== false)
        {
            $id = $fileop[0];
            $name = $fileop[1];
            $sql = "INSERT INTO csv (id, name) VALUES ('$id', '$name')";
        }

        if ($sql)
        {
            echo "OK";
        }
}
?>

<html>
<body>
 <input type="file" name="file" id="file"> 
 <input type="submit" value="Upload" name="submit">
</body>
</html>

【问题讨论】:

  • 没有mysqli_query 任何地方...

标签: php mysql


【解决方案1】:

你忘记查询了:

$sql = mysqli_query($connection, "INSERT INTO csv (id, name) VALUES ('$id', '$name')");

【讨论】:

    【解决方案2】:

    添加 mysqli_query 并将您的 if condition 放入 while 循环中...。我在您的代码中看不到任何 form 标记???我在这里添加了post method

       ?php
        $servername = "localhost";
        $username = "root";
        $password = "";
        $database = "myexcel";
    
        // Create connection
        $connection = mysqli_connect($servername, $username, $password, $database);
    
    
        if (isset($_POST['submit']))
        {
                $file = $_FILES['file']['tmp_name'];
                $handle = fopen( $file, "r");
                while(($fileop = fgetcsv($handle, 3000, ",")) !== false)
                {
                    $id = $fileop[0];
                    $name = $fileop[1];
                    $sql = mysqli_query($connection, "INSERT INTO csv (id, name) VALUES ('$id', '$name')");
    
                if ($sql)
                {
                    echo "Ok. Inserted <br>";
                }
                }
    
    
        }
        ?>
    
        <html>
        <body>
          <form method="POST"> // add this
         <input type="file" name="file" id="file"> 
         <input type="submit" value="Upload" name="submit">
        </form>
        </body>
        </html>
    

    【讨论】:

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