PHP和Mysql返回空白页面的问题

时间:2015-01-29 10:41:49

标签: php mysql

我有一个简单的表单,我正在尝试编写代码以获取从drupal到linux重新编码php页面的句柄。我不认为我可以访问所有旧代码,但我会尽力而为。 这是HTML代码:

<form accept-charset="UTF-8" method="post" action="keyset_checkin.php">
<div>
 <label for="edit-keysetnumber">Serial Number: </label>
 <input type="text" maxlength="128" name="keysetnumber"  size="60" value="" />
</div>
<input type="submit" name="op" value="Check In Keyset"  />
</form>

然后是PHP:

<?php

$dbhost = '*********';
$dbuser = '*********';
$dbpass = '*********';
$dbname = '*********';

$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}
mysql_select_db($dbname);

/*$content .=  backLink();
    $content .= "<table border='0'><tr><td colspan='3'><strong>Keylog Report: All Keyset/Keys</strong></td></tr>";

    $content .= "</table>";
        return $content;
*/
        $content .= "<table border='0'><tr><td colspan='3'><strong>Keylog Report: All Keyset/Keys</strong></td></tr>";

    $query = "SELECT keysetnumber,keyring_id,building FROM {keyinv_key_sets} WHERE building NOT LIKE '%ZONE%' and building NOT LIKE '%SUPV%' GROUP BY keyring_id ORDER BY building ASC";

    $result = db_query($query);

    if(db_affected_rows() !=0)
    {   
        while($row = db_fetch_object($result))
        {
            $content .= "<tr><td valign='top'><font size='-3'><strong>$row->building</strong></font></td><td valign='top' width='50' align='center'><font size='-3'><strong>$row->keyring_id</strong></font></td>";

            $content .= "<td valign='top'>";

            //drupal_set_message($row->keysetnumber);   

            $query2 = "SELECT keytext, keydescription FROM {keyinv_keyset_keys} WHERE keysetnumber='$row->keysetnumber'";
            $result2 = db_query($query2);

            //drupal_set_message("Made it here");

            while($row2 = db_fetch_object($result2))
            {
                //drupal_set_message("Got Here");
                $content .= "<font size='-3'>$row2->keytext <font color='#CCCCCC'>($row2->keydescription)</font><font> ";
            }   

            $content .= "</td></tr>";
        }
    }
    else
    {
        $content .= "<tr><td>No Results</td></tr>";
    }

    $content .= "</table>";


        return $content;
?>

我已经尝试将其简化为

$content .= "<table border='0'><tr><td colspan='3'><strong>Keylog Report: All Keyset/Keys</strong></td></tr>";
$content .= "<tr><td>No Results</td></tr>";
$content .= "</table>";

它只返回一个空白页面,我认为至少这3行会返回一个带有标题的空表。我以为它应该在我所在的页面返回,表格在表格的位置。我在这做错了什么

3 个答案:

答案 0 :(得分:2)

尝试回显结果($ content)而不返回结果。

答案 1 :(得分:0)

似乎未启用错误报告。

error_reporting(E_ALL);
ini_set('display_errors', 'On');

您可以在上面的代码中使用,也可以在php.ini

中全局启用它

答案 2 :(得分:0)

尝试 -

$content = "<table border='0'><tr><td colspan='3'><strong>Keylog Report: All Keyset/Keys</strong></td></tr>";

//rest of the code

echo $content;