使用两个变量连接URL

问题描述：

What I am doing is getting latitude and longitude from databases and displaying it in $row['latitude'] format. Now what I want to t to do is which ever hyperlink user clicks it takes him to maps.google.com and show location against that particular latitude and longitude.

please help! Here is my PHP code.

    <?php

    $location=$_POST["location"]; 
    $myArray = json_decode($location,true);
    $jsonArray = $myArray[0];
    $longitude = $jsonArray['longitude']; $adress = $jsonArray['adress'];
    $con=mysqli_connect("","","","");
   // Check connection if (mysqli_connect_errno())   {   echo "Failed to
   connect to MySQL: " . mysqli_connect_error();   }

   $sql = "INSERT INTO chokh_db. gpslocations (     latitude,   longitude ,
   datetime     ) VALUES ( '$latitude' , '$longitude' ,  NOW() );";



 if (!mysqli_query($con,$sql)) {
   die('Error: ' . mysqli_error($con)); } $result = mysqli_query($con,"SELECT * FROM      _db. gpslocations where
      latitude <> 0");

        while($row = mysqli_fetch_array($result))   {       
            $lat = $row['latitude'];    
            $long = $row['longitude']   
            echo $row['datetime'] . 
                "&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;" .          
                $row['latitude'] .
                "&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;" .
                $row['longitude'] ;      
            $url = "http://maps.google.com/?q='.$lat,$long.' "; 
            echo "<a href='$url'>View</a>";    echo "<br>";   
        }
        mysqli_close($con);

    ?>

答

I don't see where you are using the $url variable that you are setting but it looks like you have a typo in setting it.

  ///Change this
  $url = "http://maps.google.com/?q= '.$lat,$long.' "; 

  ///To this
  $url = "http://maps.google.com/?q=".$lat . "," . $long;

This should produce a url of something like..

http://maps.google.com/?q=37.3325004578,-122.03099823

答

You can simply write the url variable as below.

 $url = "http://maps.google.com/?q=$lat,$long";

Above string is enclosed in double quotation marks, therefore $lat and $long variables will replaced by their values. This is similar to writing it as

$url = "http://maps.google.com/?q=".$lat.",".$long;

使用两个变量连接URL

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