PHP在一页上显示2个表

Question

我正在使用这种语法在我的页面上显示一个 PHP 表。我现在需要在该表正上方添加第二个表，但我尝试的所有语法都会引发 500 错误。如何通过 1 个与 MSSQL 的连接运行 2 个 Select 语句并填充 2 个单独的 html 表？

    $option = array();
$option['driver'] = 'mssql';
$option['host'] = 'IP Address';
$option['user'] = 'UserName';
$option['password'] = 'Password';
$option['database'] = 'DB';
$option['prefix'] = '';
$db = JDatabase::getInstance($option);
$query = $db->getQuery(true);
$query = "SELECT name, hiredate, bday, payrate, hourlypay from HRData ORDER BY name ASC";
$db->setQuery($query);
$query = $db->loadObjectList();
if ($query) 
{
    ?>
    <table border="1">
    <thead>
    <tr>
    <th>Name </th>
    <th>Hire Date </th>
    <th>Birthday </th>
    <th>Pay Rate </th>
    <th>hourlypay </th>
    </tr>
    </thead>
    <?php
    foreach ($query as $res) 
    {
        print "<tr>";
        print "<td>" . $res->name . "</td>";
        print "<td>" . $res->hiredate . "</td>";
        print "<td>" . $res->bday . "</td>";
        print "<td>" . $res->payrate . "</td>";
        print "<td>" . $res->hourlypay . "</td>";
        print "</tr>";
    }
}

编辑
这是我正在尝试调整的语法，但我不断收到 500 错误

    $option = array();
$option['driver'] = 'mssql';
$option['host'] = 'IP Address';
$option['user'] = 'UserName';
$option['password'] = 'Password';
$option['database'] = 'DB';
$option['prefix'] = '';
$db = JDatabase::getInstance($option);
$query = $db->getQuery(true);
$query = "SELECT name, MAX(Pay) As PayYTD FROM HRINFO";
$db->setQuery($query);
$query = $db->loadObjectList();
if ($query) 
{
    ?>
    <table border="1">
    <thead>
    <tr>
    <th>Name </th>
    <th>YTD Pay </th>
    </tr>
    </thead>
    <?php
    foreach ($query as $res) 
    {
        print "<tr>";
        print "<td>" . $res->name . "</td>";
        print "<td>" . "$" . round($res->PayYTD) . "</td>";
        print "</tr>";
    }
}
<br><br><br>
//Query
$query = $db->getQuery(true);
$query = "SELECT name, hiredate, bday, payrate, hourlypay from HRData ORDER BY name ASC";
$db->setQuery($query);
$query = $db->loadObjectList();
if ($query) 
{
    ?>
    <table border="1">
    <thead>
    <tr>
    <th>Name </th>
    <th>Hire Date </th>
    <th>Birthday </th>
    <th>Pay Rate </th>
    <th>hourlypay </th>
    </tr>
    </thead>
    <?php
    foreach ($query as $res) 
    {
        print "<tr>";
        print "<td>" . $res->name . "</td>";
        print "<td>" . $res->hiredate . "</td>";
        print "<td>" . $res->bday . "</td>";
        print "<td>" . $res->payrate . "</td>";
        print "<td>" . $res->hourlypay . "</td>";
        print "</tr>";
    }
}

Answer 1

您遇到的问题是您错误地使用了调用。

$query = $db->getQuery(true);
$query = "SELECT name, MAX(Pay) As PayYTD FROM HRINFO";
$db->setQuery($query);

第一行将创建一个对象，不管是哪个。该对象将在 $query 中。

第二行会立即销毁对象并给$query赋值一个字符串（这是不正确的）。

第三行期望一个对象作为setQuery的参数，但不幸的是它是一个字符串！错误。

如果你想让它正常工作，那么你需要正确使用 $query 中的对象。

我不是 Joomla 专家，因此我将您链接到如何正确执行此操作的页面：https://docs.joomla.org/Selecting_data_using_JDatabase

Answer 2

将输出放入变量中。

$StringOut = '';
$StringOut .=     '<table border="1">
    <thead>
    <tr>
    <th>Name </th>
    <th>Hire Date </th>
    <th>Birthday </th>
    <th>Pay Rate </th>
    <th>hourlypay </th>
    </tr>
    </thead>';

    foreach ($query as $res) 
    {
        $StringOut .=  "<tr>";
        $StringOut .=  "<td>" . $res->name . "</td>";
        $StringOut .=  "<td>" . $res->hiredate . "</td>";
        $StringOut .=  "<td>" . $res->bday . "</td>";
        $StringOut .=  "<td>" . $res->payrate . "</td>";
        $StringOut .=  "<td>" . $res->hourlypay . "</td>";
        $StringOut .=  "</tr>";
        }
$StringOut .=     '</table>';
#Do other logic to retrieve first table. 
#You could put it in a different variable if you like. And print when and whereever you wish.

echo $StringOut;
#Optionally close connection, done.