Question

我有一个方法，需要多少项目＆amp;从限制开始并返回值

public function comment_recent_comments_block($numbers = 10, $start_from = 1) {

    $sql = " SELECT comment.comment_id, comment.subject,
             content.content_id, content.content_type_id
                    FROM comment
                    LEFT JOIN content ON content.content_id = comment.reference_id 
                    WHERE comment.reference_table = 'content'
                    AND comment.subject IS NOT NULL AND comment.subject !='' ";
    $sql .=" ORDER BY comment.weightage DESC, comment.creation_date DESC ";
    $sql .= " LIMIT :lmt_start_from , :lmt_no_of_records  ";

......其余代码...... }

我希望PHP考虑10＆amp; 1作为默认值但不起作用。我的sql抛出错误，显示'NULL'值的错误。现在，如果我明确检查参数并设置值，如

$numbers = empty($numbers) ? 10 : $numbers;
$start_from = empty($start_from) ? 1 : $start_from;

在方法中然后它可以工作。

可以解释一下我在第一种方法中缺少的东西。为什么PHP不考虑默认值...我尝试输入字符串'10'＆amp; '1'，但那也不起作用

由于

---------------------------------------完整代码------ ---------------------------

 public function findBySql($sql, $value_a = '') {
    global $dbc;
    $stmt = $dbc->connection->prepare(" $sql ");
    if (!empty($value_a)) {
     foreach ($value_a as $key => $value) {
        if (!empty($value)) {
         if ($key == 'lmt_no_of_records' || $key == 'lmt_start_from') {
            $stmt->bindValue(":$key", $value, PDO::PARAM_INT);
         } else {
            $stmt->bindValue(":$key", $value);
         }
        } else {
         $stmt->bindValue(":$key", NULL);
        }
     }
    }
    try {
     $stmt->execute();
     $result = $stmt->fetchAll(PDO::FETCH_CLASS);
     return $result;
    } catch (PDOException $e) {
     echo "Record couldnt be fetched !!: " . $e->getMessage();
     return 0;
    }
 }



public function comment_recent_comments_block($numbers = 10, $start_from = 1) {
    $numbers = empty($numbers) ? 10 : $numbers;
    $start_from = empty($start_from) ? 1 : $start_from;
    $sql = " SELECT comment.comment_id, comment.subject,
             content.content_id, content.content_type_id
                    FROM comment
                    LEFT JOIN content ON content.content_id = comment.reference_id 
                    WHERE comment.reference_table = 'content'
                    AND comment.subject IS NOT NULL AND comment.subject !='' ";
    $sql .=" ORDER BY comment.weightage DESC, comment.creation_date DESC ";
    $sql .= " LIMIT :lmt_start_from , :lmt_no_of_records  ";
    $value_a = ['lmt_no_of_records' => $numbers, 'lmt_start_from' => $start_from ];

    $result = $this->findBySql($sql, $value_a);
        $comment_string = '';
    if (count($result) > 0) {
     $comment_string .= '<ul class="documentation_list comment">';
     foreach ($result as $records) {
        $comment_string .= '<li class="comment_subject">';
        $comment_string .= '<a href="' . HOME_URL . 'content.php?mode=2&' . 'content_id=' . $records->content_id .
                        '&content_type_id=' . $records->content_type_id . '#comment_id_' . $records->comment_id . '">';
        $comment_string .= $records->subject . '</a>';
        $comment_string .= '</li>';
     }
     $comment_string .= '</ul>';
    }

    return $comment_string;
 }

现在这个工作但如果我删除下面两个语句 $ numbers =空（$ numbers）？ 10：$ number; $ start_from = empty（$ start_from）？ 1：$ start_from; 它不会工作

Answer 1

您发布的代码中没有任何特定的PDO，我认为问题与此无关（但可能）。（
有可能）请尝试类似

的内容

<?php
class Foo {
    public function comment_recent_comments_block($numbers = 10, $start_from = 1) {
            // put the following three lines of code at the beginning of your method comment_recent_comments_block
        var_export(debug_backtrace(0, 1));
        var_export( array($numbers, $start_from) );
        die;
    }
}

$foo = new Foo;
$foo->comment_recent_comments_block();

我的输出是

array (
  0 => 
  array (
    'file' => 'something.php',
    'line' => 11,
    'function' => 'comment_recent_comments_block',
    'class' => 'Foo',
    'type' => '->',
    'args' => 
    array (
    ),
  ),
)array (
  0 => 10,
  1 => 1,
)

表示该方法实际上是在没有参数的情况下调用的，并且相应的默认值已分配给$ numbers和$ start_from。
你的输出是什么？

PDO LIMIT具有默认功能值

1 个答案: