Question

我在我的网站上创建了一个类似按钮（使用ExpressionEngine），它可以正常工作。然而，当我在另一个页面上放置几乎完全相同的代码时，类似按钮的文本总是应该放在最后一页上（与应该是什么相反，虽然刷新页面说得对，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);
    }
}
?>

代码没有：

<?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);
    }
}
?>

正如您所看到的，两段代码之间的唯一区别是$ mapID和查询。但由于某种原因，一个人工作而另一个人没有。任何人都知道可能会发生什么？我认为这是一个php问题，而不是ExpressionEngine问题。

Answer 1

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$liked的当前值设置按钮的状态，但如果设置了$_POST['like']，则您正在更改系统的状态，现在该按钮不正确。如果您切换代码并首先测试$_POST['like'] ，那么在对其进行操作后设置$like的值，您的按钮将显示正确的状态。

Answer 2

现在想出来，并且还意识到第一位代码与第二位代码有相同的问题。我做的是我将POST脚本设置为按钮状态，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";
    }
}

为什么这个页面会记住最后一页的PHP变量？

2 个答案: