为什么这个页面会记住上一页的PHP变量


Why is this page remembering PHP variables from the last page?

我在我的网站上做了一个赞按钮(它使用ExpressionEngine),它可以工作。然而,当我把几乎相同的代码放在另一个页面上时,类似按钮的文本总是它在最后一页上应该是的(与它应该是的相反,尽管刷新页面会使它正确,所以我不能只是翻转值

<?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);
    }
}
?>

正如您所看到的,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。有人知道可能发生了什么吗?我认为这更像是一个php问题,而不是ExpressionEngine问题。

我认为,真正的问题是为什么第一个例子有效。。。

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'],则表示您正在更改系统的状态,此时按钮不正确。如果切换代码并首先测试$_POST['like'],然后在对其进行操作后设置$like的值,则按钮将显示正确的状态。

现在已经弄清楚了,还意识到第一位代码和第二位代码有同样的问题。我所做的是让POST脚本设置按钮状态,就像这样:

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";
    }
}