是否有可能将php返回给js变量?例:
<script type='text/javascript'>
var h = <?php include'dbconnect.php';
$charl = (some number from another sql query)
$sql=mysql_query("selct type from locations where id='$charl'");
echo $sql; ?> ;
if(h == "hostile") {
(run some other js function)
}
</script>
我需要做的是从charl(字符位置)中获得单个文本值(类型)并将其分配给java脚本变量并在其上运行if语句。有提示吗?
这是我的代码更新。它没有返回任何错误，但它没有按我想要的方式输出。它应该只返回[类型]，它应该等于敌对、城市、农场等。它不会运行，除非整个字符串在同一行。我相信它返回整个字符串，而不仅仅是echo(我需要它)
function check_hostile() { var h = '<?php session_start(); include"dbconnect.php"; $charid=$_SESSION[''char_id'']; $charloc=mysql_fetch_array(mysql_query("select location from characters where id=''$charid''")); $charl=$charloc[''location'']; $newloc=mysql_fetch_array(mysql_query("select type from locations where id=''$charl''")); echo $newl[''type'']; ?>';
if(h == "hostile") {
if(Math.random()*11 > 8) {
find_creature();
}
}
$("#console").scrollTop($("#console")[0].scrollHeight);
}
下面是alert函数运行theis时的输出。
<?php session_start(); include"dbconnect.php"; $charid=$_SESSION['char_id']; $charloc=mysql_fetch_array(mysql_query("select location from characters where id='$charid'")); $charl=$charloc['location']; $newloc=mysql_fetch_array(mysql_query("select type from locations where id='$charl'")); print $newloc['type']; ?>
改成这个
var h = <?php include "dbconnect.php";
$charl = (some number from another sql query)
$sql=mysql_query("selct type from locations where id=$charl");
$row = mysql_fetch_row($sql);
echo json_encode($row["type"]); ?>;
json_encode()将把PHP值转换为有效的Javascript表示，您可以将其注入到脚本中。
是的，这是可能的，而且是相当普遍的做法。
但是你的代码有一个小问题，它返回一个字符串，所以你必须在javascript中把它括起来。
我已经更新了你的代码来修复这个小问题并提高代码的可读性:
<?php
include'dbconnect.php';
$charl = (some number from another sql query)
$sql=mysql_query("select type from locations where id='$charl'");
if (mysql_num_rows($sql) > 0) {
$row = mysql_fetch_array($sql);
$h = $row['type'];
} else {
$h = null;
}
?>
<script type='text/javascript'>
var h = '<?php echo $h; ?>';
if(h == "hostile") {
(run some other js function)
}
</script>