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使用Sql查询在php中输出输出

[英]Print output in php using Sql queries

I am working on PHP where i want to fetch the count of a particular column 'uid' from users table. 我在PHP上工作,我想从用户表中获取特定列“ uid”的计数。 The output window is not displaying anything. 输出窗口未显示任何内容。 following is the code. 以下是代码。 Can anybody, help me in rectifying this code. 谁能帮我纠正此代码。

    <?php
    $link = mysql_connect('localhost', 'username', 'password');
    if (!$link) {
        die('Could not connect: ' . mysql_error());
    }
    mysql_select_db("rth_db");
    $data = mysql_query("SELECT COUNT(uid) AS Total FROM users", $link);
    $number = mysql_fetch_array($data);
    echo $number;
    echo 'Connected successfully';
    mysql_close($link);
    ?>

you should address your data differently 您应该以不同的方式处理数据

echo $number["Total"];

Also, mysql is deprecated, have a look at PDO. 另外,不建议使用mysql,请看一下PDO。 This also is better for security and so on. 这对于安全性等也更好。 You can check the PHP.NET page for more info on this 您可以检查PHP.NET页面以获取更多有关此的信息

change this 改变这个

echo $number;

to

print_r($number);

OR 要么

echo $number['Total'];

Because $number is array not string . 因为$numberarray而不是string

try using this : 尝试使用这个:

<?php
    $link = mysql_connect('localhost', 'username', 'password');
    if (!$link) {
        die('Could not connect: ' . mysql_error());
    }
    mysql_select_db("rth_db");
    $data = mysql_query("SELECT COUNT(uid) AS Total FROM users", $link);
    $number = mysql_fetch_array($data);
    echo $number['total'];
    echo 'Connected successfully';
    mysql_close($link);
    ?>

i have just edited one line if u notice...hope this works :) 我刚刚编辑了一行,如果你注意到...希望这行得通:)

你也可以使用

var_dump($number);

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