Question

所以我有这个质朴和基本的PHP将检查我的数据库中的值，如果它的更正它将拉出该行的其余部分我猜但我的问题是如何回应certaine HTML如果结果是好的，如何如果否则回复其他一些html。

btw atm else无效，换句话说，如果您输入的代码不在我的数据库中，则不会显示任何内容

<?php
$db_hostname = 'localhost';
$db_database = 'codedb';
$db_username = 'root';
$db_password = '';
$table = 'users';
$field = 'code';
$test = 'first_name';


// Connect to server.
$connection = mysql_connect($db_hostname, $db_username, $db_password) OR DIE ("Unable to 
connect to database! Please try again later.");

// Select the database.
mysql_select_db($db_database)
    or die("Unable to select database: " . mysql_error());

$query = "SELECT * FROM $table WHERE $field = '{$_GET["qcode"]}'";
    $result = mysql_query($query);
    if ($result) {
        while($row = mysql_fetch_array($result)) {
                $name = $row["$field"];
                $test = $row["$test"];
                echo "Hello: $name $test";
            }
    }
    else {
        echo "Im sorry you buddy, you are not a winner this time! $test";
    }
   mysql_close($connection);
?>

Answer 1

您可以使用mysql_num_rows()，这是最好的方式。

if( mysql_num_rows($result) > 0 ){
/* Anything you want here on success */
} else {
/* Anything you want here on failure */
}

Answer 2

对查询使用mysql_num_rows。

if(mysql_num_rows($result)){ } else { }

Answer 3

您可以简单地将html包含在if语句

中，而不是回显每个html值

像：

<?php
$query = "SELECT * FROM $table WHERE $field = '{$_GET["qcode"]}'";
$result = mysql_query($query);
if ($result) {
    while($row = mysql_fetch_array($result)) {
            $name = $row["$field"];
            $test = $row["$test"];
?>
<b>Hello: <?php echo $name $test"; ?></b>
<?php
        }
}
else {
?>
    <b> Im sorry you buddy, you are not a winner this time! </b> <?php echo $test"; 
}
mysql_close($connection);
?>

Answer 4

if(mysql_num_rows($result) > 0) { echo "output"; } else { echo "error msg"; }

Answer 5

首先，您必须更正这样的数据库选择语法

mysql_select_db($db_database,$connection);

Insted of

mysql_select_db($db_database);

检查Manual

mysql_query如何显示if条件的html

5 个答案: