Question

我正在尝试从mysql获取数据并使用while循环显示它们。但是问题出在while循环内，我得到的数据总是少一个。假设我的数据库中有两行，但是使用此代码，我只会得到一行。第一行总是丢失。弄清楚为什么！共享一些代码。

尝试var_dump（），它显示db中有正确的数字行

 $ddaa = mysql_query("SELECT * FROM coupons ORDER BY id");
    echo mysql_error();
  $data = mysql_fetch_array($ddaa);
while ($data = mysql_fetch_array($ddaa))
  {
echo $data['id'] ;
}

Answer 1

您正在使用while循环之前获取一行，而您不在任何地方使用while循环，这就是您丢失一行的原因。

$ddaa = mysql_query("SELECT * FROM coupons ORDER BY id") or die(mysql_error());

while ($data = mysql_fetch_array($ddaa))
{
      echo $data['id'] ;
}

Answer 2

尝试删除此行：

$data = mysql_fetch_array($ddaa);

Answer 3

您的代码中缺少服务器和数据库凭据，请尝试执行此操作

$server = 'server_name';
$user = 'server_username';
$pass = 'server_password';
$db = 'database_name';

$connection = new mysqli($server, $user, $pass, $db);

$aa = "SELECT * FROM coupons ORDER BY id"; 
$dd = mysqli_query($connection,$aa); // $connection is the variable which contains server  and database  credentials;
while ($data = mysqli_fetch_assoc($dd)) {
    echo $data['id'];
}

Answer 4

它将为我工作。试试这个...

<?php

    $con=mysql_connect('localhost','root','') or die("could not connect".mysql_error());

    mysql_select_db('dbname');
    $query = mysql_query("SELECT * FROM Student");
    $num_rows = mysql_num_rows($query);
    while($row = mysql_fetch_array($query))
    {
        echo $row['firstname'];
    }

    echo "<h3>Record Selected successfully\n</h3>";
    mysql_close($con);

    ?>

而（$ data = mysql_fetch_array（））不起作用

4 个答案: