Question

我想通过从我的数据库中获取纬度，经度以及查询（SELECT * FROM trackEmployee）它来检索地图中的点数数据并在地图上显示（Y）

但是当我试图通过在帖子中获取数据来设置条件WHERE groupEmail ='$mail' AND date= '$date'时它不起作用

如果我直接将值传递给条件，则它起作用（Y）

这里的问题是从上面的php标签获取包含查询的下一个php标签。

<html>
 <head>
 <meta http-equiv="content-type" content="text/html; charset=utf-8"/>
 <title>Google Map API V3 with markers</title>
 <style type="text/css">
 body { font: normal 10pt Helvetica, Arial; }
 #map { width: 750px; height: 400px; border: 0px; padding: 0px; }
 </style>

<?php


$connect_mysql= @mysql_connect($server,$username,$passwor) or die ("Connection Failed!");
$mysql_db=mysql_select_db("GP15",$connect_mysql) or die ("Could not Connect to Database");

$mail=$_POST['sel1'];
echo $mail;  // prints correctly 
$date=$_POST['sel2'];
echo $date;  // prints correctly 

//On page 2
?>

 <script src="http://maps.googleapis.com/maps/api/js?sensor=false" type="text/javascript"></script>
 <script type="text/javascript">
 //Sample code written by August Li
 var icon = new google.maps.MarkerImage("http://maps.google.com/mapfiles/ms/micons/blue.png",
 new google.maps.Size(32, 32), new google.maps.Point(0, 0),
 new google.maps.Point(16, 32));
 var center = null;
 var map = null;
 var currentPopup;
 var bounds = new google.maps.LatLngBounds();



 function addMarker(lat, lng, info) {
 var pt = new google.maps.LatLng(lat, lng);
 bounds.extend(pt);
 var marker = new google.maps.Marker({
 position: pt,
 icon: icon,
 map: map
 });
 var popup = new google.maps.InfoWindow({
 content: info,
 maxWidth: 300
 });
 google.maps.event.addListener(marker, "click", function() {
 if (currentPopup != null) {
 currentPopup.close();
 currentPopup = null;
 }
 popup.open(map, marker);
 currentPopup = popup;
 });
 google.maps.event.addListener(popup, "closeclick", function() {
 map.panTo(center);
 currentPopup = null;
 });
 }
 function initMap() {
 map = new google.maps.Map(document.getElementById("map"), {
 center: new google.maps.LatLng(0, 0),
 zoom: 14,
 mapTypeId: google.maps.MapTypeId.ROADMAP,
 mapTypeControl: false,
 mapTypeControlOptions: {
 style: google.maps.MapTypeControlStyle.HORIZONTAL_BAR
 },
 navigationControl: true,
 navigationControlOptions: {
 style: google.maps.NavigationControlStyle.SMALL
 }
 });

  <?php

// $ query = mysql_query（＆＃34; SELECT * FROM trackEmployee＆＃34;）; //此查询有效

 $query = mysql_query("SELECT * FROM trackEmployee WHERE employeeEmail='$mail'AND date='$date'");


 while ($row = mysql_fetch_array($query)){


 $name=$row['street'];
 $lat=$row['latitude'];
 $lon=$row['longitude'];
 $desc=$row['district'];

 echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc')\n");

}

 ?>
 center = bounds.getCenter();
 map.fitBounds(bounds);

 }
 </script>



 </head>
 <body onload="initMap()" style="margin:0px; border:0px; padding:0px;">


 <div id="map"></div>
 </html>

Answer 1

用this替换你的查询。希望它有效。

$query = mysql_query("SELECT * FROM trackEmployee WHERE employeeEmail='".$mail."'AND date='".$date."'");

将变量从php标记传递到同一页面中的其他标记

1 个答案: