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我在我的网站（使用 ExpressionEngine）上做了一个点赞按钮，它可以工作。但是，当我将几乎相同的代码放在另一个页面上时，like 按钮的文本始终是它应该在最后一页上的文本（与它应该的相反，尽管刷新页面会正确，所以我不能只是翻转值。这是有效的代码：

<?php
$DB1 = $this->EE->load->database('ext_db', TRUE);
$mapID = "{entry_id}";
$ipAddress = $_SERVER['REMOTE_ADDR'];
$thisquery = "SELECT * FROM mapLikes WHERE ipAddress = '$ipAddress' AND mapID = '$mapID'";
$q = $DB1->query($thisquery);
$results = $q->result_array();
foreach ($q->result() as $row)
{
    $liked = $row->liked;
}
$buttontext = 'Like';
$buttonimage = "1";
if ($liked == "1") {
    $buttontext = 'Unlike';
    $buttonimage = "2";
}

if($_POST['like']) {
    if ($liked == "1") {
        $thisquery = "DELETE FROM mapLikes WHERE mapID = '$mapID' AND ipAddress = '$ipAddress'";
        $DB1->query($thisquery);
    } else {
        $thisquery = "INSERT INTO mapLikes (mapLikeID, mapID, liked, ipAddress) VALUES ('null', '$mapID', '1', '$ipAddress')";
        $DB1->query($thisquery);
    }
}
?>

而没有的代码：

<?php
$DB1 = $this->EE->load->database('ext_db', TRUE);
$mapID = "{segment_3_category_id}";
$ipAddress = $_SERVER['REMOTE_ADDR'];
$thisquery = "SELECT * FROM categoryLikes WHERE ipAddress = '$ipAddress' AND mapID = '$mapID'";
$q = $DB1->query($thisquery);
$results = $q->result_array();
foreach ($q->result() as $row)
{
    $liked = $row->liked;
}
$buttontext = 'Like';
$buttonimage = "1";
if ($liked == "1") {
    $buttontext = 'Unlike';
    $buttonimage = "2";
}

if($_POST['like']) {
    if ($liked == "1") {
        $thisquery = "DELETE FROM categoryLikes WHERE mapID = '$mapID' AND ipAddress = '$ipAddress'";
        $DB1->query($thisquery);
    } else {
        $thisquery = "INSERT INTO categoryLikes (categoryLikeID, mapID, liked, ipAddress) VALUES ('null', '$mapID', '1', '$ipAddress')";
        $DB1->query($thisquery);
    }
}
?>

如您所见，这两段代码之间的唯一区别是 $mapID 和查询。然而由于某种原因，一个有效，另一个无效。有人知道会发生什么吗？我认为这更像是一个 php 问题，而不是 ExpressionEngine 问题。

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2 回答 2

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我认为，真正的问题是为什么第一个例子确实有效......

您正在根据 的当前值设置按钮的状态$liked，但是如果$_POST['like']设置了，则您正在更改系统的状态并且按钮现在不正确。如果您切换代码并测试$_POST['like'] first，然后在您对其进行操作后设置 的值$like，您的按钮将显示正确的状态。

于 2013-08-08T22:55:14.123 回答
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现在想通了，也意识到第一段代码和第二段代码有同样的问题。我所做的是让 POST 脚本设置按钮状态，如下所示：

if($_POST['like']) {
    if ($liked == "1") {
        $thisquery = "DELETE FROM mapLikes WHERE mapID = '$mapID' AND ipAddress = '$ipAddress'";
        $DB1->query($thisquery);
        $buttontext = 'Like';
        $buttonimage = "1";
    } else {
        $thisquery = "INSERT INTO mapLikes (mapLikeID, mapID, liked, ipAddress) VALUES ('null', '$mapID', '1', '$ipAddress')";
        $DB1->query($thisquery);
        $buttontext = 'Unlike';
        $buttonimage = "2";
    }
}
于 2013-08-14T09:58:26.960 回答