Question

我对php比较新。我有一个函数来通过tableName获取所有主题。我也有一个函数，我试图获取subject_id（来自pages表）匹配主题表id的所有页面。

以下是错误和输出：

PDOStatement对象（[queryString] =＆gt; SELECT * FROM subject ORDER BY id ASC）

数组（[0] =＆gt;数组（[id] =＆gt; 1 [0] =＆gt; 1 [menu_name] =＆gt;关于我们[1] =＆gt;关于我们[position] =＆gt; 1 [2] =＆gt; 1 [可见] =＆gt; 1 [3] =＆gt; 1）[1] =＆gt;数组（[id] =＆gt; 2 [0] =＆gt; 2 [menu_name] =＆gt;产品[1] =＆gt;产品[位置] =＆gt; 2等等......

注意：第39行/Users/aaronhappe/Sites/php/CMS/includes/db.php中的未定义索引：id

致命错误：在第41行的/Users/aaronhappe/Sites/php/CMS/includes/db.php中的非对象上调用成员函数rowCount（）

以下是代码：

<?php

// controller file

include '../admin-head-includes.php';

$subjects = getAll('subjects', $conn);

$subjectId = $subjects->fetchAll();

print_r($subjects);
echo "<br>";
echo "<br>";
print_r($subjectId);

$pages = getAllLinkedId('pages', $subjectId, $conn);

view('manage-content', array(
    'subjects' => $subjects,
    'pages'    => $pages,
));

// from my db functions file

function getAllLinkedId($tableName, $tableName2, $conn) {

try {
$result = $conn->query("SELECT * FROM $tableName WHERE subject_id = {$tableName2['id']} ORDER BY id   ASC");

    return ($result->rowCount() > 0)
    ? $result
    : false;

} catch (Exception $e) {
    return false;
}

<!-- standard functions file -->

<?php function view($path, $data = null) {
    if ($data) {
        extract($data);
    }
    $path = $path . '.views.php';
    include "views/$path";
}?>

如果我将$ subject [0]传递给我的getAllLinkedId函数，它解决了Array包装其他数组的问题。但是，我想要的是为每个主题吐出相应的页面。不仅仅是对应于一个单一主题的页面，而且它对所有主题作出响应。

这是我的观点：

<div class="subjects">
    <ul>
<?php

foreach ($subjects as $subject) {?>
<li>
<?=$subject['menu_name'];?>
</li>
    <ul>
    <?php foreach ($pages as $page) {?>
        <?=$page['menu_name']?>
    <?php }?>
    </ul>

<?php }?>
</ul>

</div>

Answer 1

第一个问题：

Notice: Undefined index: id in ...

未设置数组$tableName2中的索引ID，请检查是否已设置

第二个问题：

Fatal error: Call to a member function rowCount() on a non-object ..

$conn->query() on失败返回非对象，检查$result是否为对象

在您的代码中：

function getAllLinkedId($tableName, $tableName2, $conn) {

    // check if index id is set
    if( !isset($tableName2['id']) )
        return false;

    try {
        $result = $conn->query("SELECT * FROM $tableName WHERE subject_id = {$tableName2['id']} ORDER BY id ASC");

        // check if $result is object
        if( !is_object($result) )
            return false;

        return ($result->rowCount() > 0) ? $result : false;

    } catch (Exception $e) {
        return false;
    }
}

Answer 2

我离开了这个基础。但我已经弄清楚了。我发布我的解决方案来关闭它。

<?php

// controller file

include '../admin-head-includes.php';

$subjects = getAll('subjects', $conn);
$pages    = getAll('pages', $conn);

$subjects = $subjects->fetchAll();
$pages    = $pages->fetchAll();

view('manage-content', array(
    'subjects' => $subjects,
    'pages'    => $pages,
));

<!-- what my view function does -->

<?php function view($path, $data = null) {
if ($data) {
    extract($data);
}
$path = $path . '.views.php';
include "views/$path";
}?>

<!-- view -->
<div class="subjects">
<ul>
<?php

foreach ($subjects as $subject) {?>
<li>
<?=$subject['menu_name'];?>
<ul>
<li>
<?php foreach ($pages as $page) {?>
<ul>
    <li>
<?php
if ($page['subject_id'] == $subject['id']) {
    echo $page['menu_name'];
}
    ?>
</li>
</ul>
<?php }?>
</li>
</ul>

<?php }?>
</li>
</div>
</ul>

PHP匹配主题表和页表

2 个答案: