mysql - Joomla数据库查询3.2

Question

您好，我正在尝试进行查询，使登录用户可以获得 pokalista 表 3 行的值。无法正确帮助任何人吗？

?php

$user = JFactory::getUser();
if ($user->guest) 
{ 
    echo "<h2>Login first.</h2>"; 
}
else
{
    $db = JFactory::getDbo();
    $query = $db->getQuery(true);
    $query->select('*');
    $query->from($db->quoteName('fab_pokalista'));
    // it should pickup current login user username and return 3th value of row
    // name colum has same values that users real name
    $query->where($db->quoteName('name')." = ".$db->quote('$user->name .'));
    $db->setQuery($query);
    $row = $db->loadRow();
    echo "<h3>Hello:</h3><br />";
    echo $user->name . '<br />';
    echo "<h3>value of query:</h3>"; 
    print_r($row[3]);
}
?>

Answer 1

您的 where 语句中存在引用错误以及多余的空格和句点，请参阅下一行中的粗体部分:

$query->where($db->quoteName('name')." = ".$db->quote( '$user->name .' ));

You're using single quotes around the variable reference $user->name if you want PHP to convert that to a value you need to either use double quotes " or you can remove them altogether.

With double quotes

$query->where($db->quoteName('name')." = ".$db->quote("$user->name"));

不带引号

    $query->where($db->quoteName('name') . " = " . $db->quote($user->name));


<?php
$user = JFactory::getUser();

if ($user->guest)
{
    echo "<h2>Login first.</h2>";
}
else
{
    $db = JFactory::getDbo();
    $query = $db->getQuery(true);
    $query->select('*');
    $query->from($db->quoteName('fab_pokalista'));
    // it should pickup current login user username and return 3th value of row
    // name colum has same values that users real name
    $query->where($db->quoteName('name') . " = " . $db->quote($user->name));
    $db->setQuery($query);
    $row = $db->loadRow();
    echo "<h3>Hello:</h3><br />";
    echo $user->name . '<br />';
    echo "<h3>value of query:</h3>";
    print_r($row[3]);
}