【问题标题】:How can i call a php variable from a seperate file into a HTML file如何将单独文件中的 php 变量调用到 HTML 文件中
【发布时间】:2014-02-10 08:13:58
【问题描述】:

这是我想从中分离 javascript 的实际文件。 我如何告诉 javascript 在哪里寻找 php 文件,目前我只能从 php 文件中执行 javascript..

<?php
date_default_timezone_set('Europe/London');
require_once("DbConnect.php");

$sql = "SELECT `artist`, `title`, `label`, `albumyear`, `date_played`, `duration`,
`picture` FROM historylist  ORDER BY `date_played` DESC LIMIT 5 ";

$result = $db->query($sql);

$lastplayed = array();
$i = 1;
while ($row=$result->fetch_object()) {
$lastplayed[$i]['artist'] = $row->artist;
$lastplayed[$i]['title'] = $row->title;
$lastplayed[$i]['label'] = $row->label;
$lastplayed[$i]['albumyear'] = $row->albumyear;
$lastplayed[$i]['date_played'] = $row->date_played;
$lastplayed[$i]['duration'] = $row->duration;
$lastplayed[$i]['picture'] = $row->picture;
$i++;
}

$starttime = strtotime($lastplayed[1]['date_played']);
$curtime = time();
$timeleft = $starttime+round($lastplayed[1]['duration']/1000)-$curtime;
$secsremain = (round($lastplayed[1]['duration'] / 1000)-($curtime-$starttime));


$lastplayedjson = json_encode( $lastplayed );
?>

我希望这部分放在单独的文件中

<script type="text/javascript">


var lastplayedjson = <?php echo $lastplayedjson ?>;


alert( lastplayedjson[1].title );
alert( lastplayedjson[2].title );
alert( lastplayedjson[3].title );
alert( lastplayedjson[4].title );
</script>

【问题讨论】:

  • (1) 什么时候需要检查变量? (2) $lastplayed 什么时候更新?

标签: javascript php json


【解决方案1】:

你可以include第一个php文件(我们称之为fileA.php),它允许fileB.php访问它的所有变量

文件 A

<?php 
$cats = 'topkek';

文件 B

<?php
include('/path/to/fileA.php');
echo $cats;

应该输出

topkek

您可以查看here 了解includerequireinclude_oncerequire_once 上的差异

【讨论】:

    【解决方案2】:

    在你的 php 中:

    include_once('templates/template-name.php');
    

    【讨论】:

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