插入php代码后,映射页面变为空白

插入php代码后,映射页面变为空白,php,google-maps-api-3,maps,infowindow,Php,Google Maps Api 3,Maps,Infowindow,将php代码插入infowindow内容后,我的gmap页面变为空白 google.maps.event.addListener(noi, 'mouseover', function() { var infowindow = new google.maps.InfoWindow({ content: '<?php if ($count==0){ echo "No Open Tickets"; } else{

将php代码插入infowindow内容后,我的gmap页面变为空白

google.maps.event.addListener(noi, 'mouseover', function() {
  var infowindow = new google.maps.InfoWindow({
    content: '<?php 
        if ($count==0){
        echo "No Open Tickets";
        }
    else{
        echo "<table>";
        foreach ($NOIcompliancearray as $SLA_Compliance=>$count) {
            $Npath = $Nimages[$SLA_Compliance];
            echo "<tr>";
            echo "<td><a href='city.php?city=Noida&compliance=".$SLA_Compliance."'><img src='IndiaImages/".$Npath."' title='".$SLA_Compliance."' ></td>";
            echo "<td>".$count."</td>";
            echo "</tr>";
            }
        echo "</table>";        
    }
  ?>'
    size: new google.maps.Size(100,100),
  });
google.maps.event.addListener(noi, 'mouseover', function() {
infowindow.open(map,noi);
setTimeout(function() { infowindow.close(map, noi) }, 5000);
});
google.maps.event.addListener(noi,'mouseover',function(){
var infowindow=new google.maps.infowindow({
内容:“查看函数

google.maps.event.addListener(noi, 'mouseover', function() {
  var infowindow = new google.maps.InfoWindow({
    content: '...'
    size: new google.maps.Size(100,100),
  });
使用php,您可以

google.maps.event.addListener(noi, 'mouseover', function() {
  var infowindow = new google.maps.InfoWindow({
    content: '<table>
            <tr>
            <td><a href='
   // 'city.php?city=......' is after content : string ending sign `'` will be ignored by function !
    size: new google.maps.Size(100,100),
  });
改为转义
符号

echo "<td><a href=\"city.php?city=Noida&ompliance=\"".$SLA_Compliance."\">"

echo“

PHP运行服务器端。infowindow代码在客户端运行。在页面上查看源代码。在“查看源代码”下,它给出以下输出:内容:“310Noida”。我是否应该编写javascript来获取显示的数据?请提出建议。@moskito-x:进行了修改,但系统停止了“分析错误:语法错误,意外的'indaimages'(T_字符串),在C:\xampp\htdocs\map\gm4v2.php的第xx行中应为','或';”。同样的代码在css层中工作没有任何问题。@TrinathKanagala:我的错误,我读错了。我删除了我的注释。但是你得到了什么:
echo”@moskito-x:很抱歉,你对代码做了一些修改。没有注意到。现在它显示了它应该显示的方式。非常感谢。另外,我想了解你为什么在中间添加斜杠。你是否可以解释或引导我到适当的频道,以便我能更好地理解?
echo "<td><a href=\"city.php?city=Noida&ompliance=\"".$SLA_Compliance."\">"