更新：尝试在PHP中存储数组，然后调用一个javascript函数

Z时代
2024-01-10
分类：问答

我试图将结果存储在使用PHP的MySQL数组中，然后使用结果调用JavaScript函数以在那里使用结果。我不知道为什么我的地图上没有显示出来（试图实现谷歌地图在我的网页）更新：尝试在PHP中存储数组，然后调用一个javascript函数

我的PHP/HTML/JavaScript调用

<?php 
..... 
<div id="content" 
      ...... 
$address=array(); 
//will list out where to go 
while ($sec = mysql_fetch_array($result2)) { 
     $address[$x++] = ($sec[5] . " " . $sec[7]); 
} 
print_r($latlng); 
print_r($address); 
mysql_close($link); 
?> 
<div id="address_container"> 
<?php 
print array_shift($address); 
?> 
</div> 
</div>

我的Javascript代码：

<script type="text/javascript"> 
var geocoder; 
    var map; 
    function initialize() { 
    geocoder = new google.maps.Geocoder(); 
    var latlng = new google.maps.LatLng(-34.397, 150.644); 
    var myOptions = { 
     zoom: 8, 
     center: latlng, 
     mapTypeId: google.maps.MapTypeId.ROADMAP 
    } 
    map = new google.maps.Map(document.getElementById("map_canvas"), myOp$ 
    } 
function codeAddress() { 
var address = document.getElementByID("address_container").innerHTML; 
console.log(address); 
    geocoder.geocode({ 'address': address}, function(results, status) { 
     if (status == google.maps.GeocoderStatus.OK) { 
     map.setCenter(results[0].geometry.location); 
     var marker = new google.maps.Marker({ 
      map: map, 
      position: results[0].geometry.location 
     }); 
     } else { 
     alert("Geocode was not successful for the following reason: " + s$ 
     } 
    }); 
}

回答：

看是否有此产生的结果你期待：

PHP：

<?php 
    // ...... 
    $address = array(); 
    while ($sec = mysql_fetch_array($result2)) $address[++$x]= "$sec[5] $sec[7]"; 
    // this should print the same data as you were getting before if you un-comment it 
    // print_r($address); 
    mysql_close($link); 
?> 
<div id="address_container"><?php print array_shift($address); ?></div> 
</div> <!-- I left this here because I don't know what the rest of your HTML looks like... --> 
<script type="text/javascript"> 
document.write(codeAddress()); 
</script>

的Javascript：

function codeAddress() { 
    var address = document.getElementByID("address_container").innerHTML; 
    geocoder.geocode({ 'address': address}, function(results, status) { 
    if (status == google.maps.GeocoderStatus.OK) { 
     map.setCenter(results[0].geometry.location); 
     var marker = new google.maps.Marker({ 
     map: map, 
     position: results[0].geometry.location 
     }); 
    } else { 
     alert("Geocode was not successful for the following reason: " + status); 
    } 
    }); 
}

回答：

不print_r的$地址，但呼应它变成一个JS变种

</div> 
<script type="text/javascript"> 
var address = <?php echo json_encode($address); ?> ; 
document.write(codeAddress()); 
</script>

那么没有必要在你的函数来选择它

回答：

这是什么例子我正要离开...从谷歌api

var geocoder; 
    var map; 
    function initialize() { 
    geocoder = new google.maps.Geocoder(); 
    var latlng = new google.maps.LatLng(-34.397, 150.644); 
    var myOptions = { 
     zoom: 8, 
     center: latlng, 
     mapTypeId: google.maps.MapTypeId.ROADMAP 
    } 
    map = new google.maps.Map(document.getElementById("map_canvas"), myOptions); 
    } 
    function codeAddress() { 
    var address = document.getElementById("address").value; 
    geocoder.geocode({ 'address': address}, function(results, status) { 
     if (status == google.maps.GeocoderStatus.OK) { 
     map.setCenter(results[0].geometry.location); 
     var marker = new google.maps.Marker({ 
      map: map, 
      position: results[0].geometry.location 
     }); 
     } else { 
     alert("Geocode was not successful for the following reason: " + status); 
     } 
    }); 
    } 
<body onload="initialize()"> 
<div id="map_canvas" style="width: 320px; height: 480px;"></div> 
    <div> 
    <input id="address" type="textbox" value="Sydney, NSW"> 
    <input type="button" value="Encode" onclick="codeAddress()"> 
    </div> 
</body>