我试图用PHP变量参数调用JavaScript函数.
我尝试了两种方法.
>使用echo中的脚本标记在PHP中调用JavaScript函数
即
<?PHP
echo '<script>initialize('.$lat.','.$lang.','.$zom.');</script>';
?>
>将PHP变量值分配给JavaScript变量
<script> var lat="<?PHP echo $lat;?>"; var lang="<?PHP echo $lang; ?>"; var zoom="<?PHP echo $zoom; ?>"; alert(lat+lang+zoom); initialize(lat,lang,zoom); </script>
在第一种情况下,函数被调用,因为我从页面源交叉检查,但传递的参数是未定义的.
在第二种情况下,值成功保存在JavaScript变量中,通过alert()进行检查,但不调用函数.
这是整个代码：
<!DOCTYPE html>
<html>
<head>
<script src="http://maps.googleapis.com/maps/api/js?key=AIzaSyDY0kkJiTPVd2U7aTOAwhc9ySH6oHxOIYM&sensor=false">
</script>
<?PHP
if( isset($_POST['lat']) && isset($_POST['lang']) && isset($_POST['zoom']) && isset($_POST['city'])):
$lat=$_POST['lat'];
$lang=$_POST['lang'];
$zoom=$_POST['zoom'];
$city=$_POST['city'];
$zom=(int)$zoom;
var_dump($lang);
var_dump($lat);
//var_dump($zoom);
var_dump($zom);
//echo '<script>initialize('.$lat.','.$lang.','.$zom.');</script>';
endif;
?>
<script>
var lat="<?PHP echo $lat; ?>";
var lang="<?PHP echo $lang; ?>";
var zoom="<?PHP echo $zoom; ?>";
alert(lat+lang+zoom);
initialize(lat,lang,zoom);
</script>
<script>
function initialize(a,b,zom){
if (!a || !b ||!zom){
alert('came on not' +a+b +zom);
// var centerLoc=new google.maps.LatLng( 33.61701054652337,73.37824736488983);
zoom=16;
}
else
{
alert('came');
var zoom =parseInt(zom);
var centerLoc=new google.maps.LatLng(a,b);
}
var mapProp = {
center:centerLoc,
zoom:zoom,
//mapTypeId:google.maps.MapTypeId.ROADMAP
mapTypeId:google.maps.MapTypeId.SATELLITE
};
var map=new google.maps.Map(document.getElementById("googleMap") ,mapProp);
marker=new google.maps.Marker({
position:centerLoc,
title:'Click to zoom'
});
google.maps.event.addListener(marker,'click',function() {
map.setZoom(map.getZoom()+1);
map.setCenter(marker.getPosition());
});
marker.setMap(map);
}
google.maps.event.addDomListener(window, 'load', initialize);
</script>
</head>
<body style= "background-color:gainsboro;">
<form method="POST" action="myPage.PHP" >
Enter latitude: <input type ="text" name="lat" id="lat" / ><br/>
Enter longitude: <input type ="text" name="lang" id="lang"/ ><br/>
Enter City Name: <input type="text" name="city" id="city"/><br/>
Enter Zoom level: <input type ="text" name="zoom" id="zoom"/ ><br/>
<input type="button" value ="Perview" onclick=" initialize(
document.getElementById('lat').value,
document.getElementById('lang').value,
document.getElementById('zoom').value)"/>
<input type="Submit" value="Save" />
</form>
<center><div id="googleMap" style="width:1000px;height:500px;"></div></center>
</body>
</html>
解决方法:
使用json_encode().如果不这样做,那么当它从PHP传递到HTML / JS层时,总是有可能错误地转义数据.
$vars = array($lat, $lang, $zoom);
// JSON_HEX_TAG and JSON_HEX_AMP are to remove all possible surprises that Could be
// caused by vars that contain '</script>' or '&' in them. The rules for
// escaping/encoding inside script elements are complex and vary depending
// on how the document is parsed.
$jsvars = json_encode($vars, JSON_HEX_TAG | JSON_HEX_AMP);
echo "<script>initialize.apply(null, $jsvars)</script>";
通常,为了您的理智,您需要为当前页面上运行的js提供的所有PHP数据应该被收集到一个PHP数组中,然后放入单个js对象中.例如：
<?PHP
$jsdata = array(
'formvars' => array(
'lat' => $lat,
'lang' => $lang,
'zoom' => $zoom
),
'username' => $username,
'some_other_data' => $more stuff
);
?>
<script>
var JSDATA = <?=json_encode($jsdata, JSON_HEX_TAG | JSON_HEX_AMP )?>;
initialize(JSDATA.formvars.lat, JSDATA.formvars.lang, JSDATA.formvars.zoom);
</script>
