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PHP MySQLi,如果ID = 1,则返回行名

[英]PHP MySQLi if ID = 1 echo name from row

我正在尝试制作仅显示ID为1的行名称的东西,但似乎无法正常工作。 我可以使它显示所有名称,但我只希望它显示用户ID 1的名称。这是我当前的代码,但是不起作用。

<a style="font-size: 17px; color: #ff0000;"><?php 
$q = "SELECT * FROM `Team` WHERE id =1";

$result=mysqli_query($q);

    $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
    if ($row != FALSE) {
        echo '<br />$row is not false.';
        $name = $row['name'];
        echo $name;
    } else{echo "it's false :(";};
    ?></a>

它返回:

这是错误的:(

您可能需要while()检查。 尝试类似:

您的数据库连接:

$servername = "YOUR_HOST";
$username = "YOUR_USER";
$password = "YOUR_PASSWORD";
$dbname = "YOUR_DATABASE"; 

$mysqli = new mysqli($servername, $username, $password, $dbname);
if ($mysqli->connect_error) {
    echo "There was a slight problem, please contact your webmaster before continuing.";
    exit();
}

然后,您的主文件将显示您想要的行:

// create query
$q = "SELECT * FROM Team WHERE id = 1";

// get the records from the database
if ($result = $mysqli->query($q))
{
    // display records if there are records to display
    if ($result->num_rows > 0)
    {
        // fetch the results
        while ($row = $result->fetch_object())
        {
            $name = $row->name;
            echo $name;
        }

    }
    else
    {
        echo "No results to display!<br><hr><br>";
    }

}
else
{ // show an error if there is an issue with the database query
    echo "Error: " . $mysqli->error;
}
// close database connection
$mysqli->close();

mysqli_query要求第一个参数应该是连接字符串,第二个是查询

mysqli_query($ link,“您的查询”);

参考: http : //php.net/manual/en/mysqli.query.php

您需要添加连接参数!

$result=mysqli_query($db, $q);

代替

$result=mysqli_query($q);

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