我可以从PHP数组中移动数组从JavaScript数据库回收 [英] can i move array from php to javascript retrive from mysql database

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问题描述

我的php代码get_marker_connect2.php

 <?php 
$ servername =localhost;
$ username =root;
$ passcode =;
$ dbname =complaint_system;
$ con = mysqli_connect($ servername,$ username,$ passcode,$ dbname);
//检查连接
if(mysqli_connect_errno()){
echo无法连接到MySQL:。 mysqli_connect_error();
}

//存储查询
$ result = mysqli_query($ con,SELECT l_longitude,l_latitude FROM register_complain);

echo< table border ='1'>

< tr>
th纬度< />
< /< / ;经度< / th>
< / tr>;

while($ row = mysqli_fetch_array($ result)){
echo< tr>;
回显< td> 。 $ row ['l_longitude']。 < / TD> 中;
回显< td>
。 $ row ['l_latitude']。 < / TD> 中;
回声< / tr>;
}
echo< / table>;
header('Location:http://localhost/cca/View_Map.html');
mysqli_close($ con);
?>

这是我的javascript,命名为javascript1.js

  function initialize(){
var myLatLng = new google.maps.LatLng(24.86,67.01);
var mapOptions = {

center:myLatLng,
zoom:14,
mapTypeControl:true,
mapTypeId:google.maps.MapTypeId.ROADMAP} ;

alert('alert array here');
$ b $ * for(var i = 0; i< latlong1.length; i ++){
alert(Longitude =。latlong1 [i] .longitude);
alert(Latitude =。latlong1 [i] .latitude);
}
* /

map = new google.maps.Map(document.getElementById(view-map),
mapOptions);
var marker = new google.maps.Marker({
position:simple,
map:map,
title:'Hello Karachi!'});
}
google.maps.event.addDomListener(window,'load',initialize);

可以直接移动php数组$ row ['l_longitude']和$ row ['l_latitude']到我的JavaScript?我有谷歌它很多,但还没有找到解决方案。
i希望在我的javascript中提醒该数组,在alert array here中

解决方案

如果您希望打印复杂结构插入到在浏览器中工作的页面中,您需要Javascript可以解析的语法,并且可以打印服务器端语言。

neareset的可能性为:


  • XML

  • JSON



我现在推荐使用JSON。


$ b 使用PHP,您可以在某处写入:

echovar myTransportedArrayJson = \。json_encode($ my_array)。\;



然后让Javascript解析你的json表达式:

  var myTransportedArray = JSON.parse(myTransportedArrayJson); 
alert(myTransportedArray); //<<<在这里,你是。

当然,片段必须是响应的一部分,否则它不会到达服务器。另外,你应该用 try .. catch 来解析。如果继续工作并使用更复杂的结构,则必须考虑转义某些字符。所以你现在会发现一些文献:-)


my php code get_marker_connect2.php

<?php
$servername = "localhost";
$username = "root";
$passcode = "";
$dbname = "complaint_system";
$con=mysqli_connect($servername,$username,$passcode,$dbname);
// Check connection
if (mysqli_connect_errno()) {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

//store query
$result = mysqli_query($con,"SELECT l_longitude,l_latitude FROM register_complain");

echo "<table border='1'>

<tr>
<th>Latitude</th>
<th>Longitude</th>
</tr>";

while($row = mysqli_fetch_array($result)) {
    echo "<tr>";
    echo "<td>" . $row['l_longitude'] . "</td>";
    echo "<td>
    " . $row['l_latitude'] . "</td>";
    echo "</tr>";
}
echo "</table>";
header('Location:http://localhost/cca/View_Map.html');
mysqli_close($con);
?> 

This is my javascript named as javascript1.js

function initialize() {    
    var myLatLng = new google.maps.LatLng(24.86, 67.01);
    var mapOptions = {

    center: myLatLng,
    zoom: 14,
    mapTypeControl: true,
    mapTypeId: google.maps.MapTypeId.ROADMAP};

    alert('alert array here');

    /*for (var i = 0; i < latlong1.length; i++) {
    alert(" Longitude = ". latlong1[i].longitude);
    alert(" Latitude = ". latlong1[i].latitude);
    }
    */

    map = new google.maps.Map(document.getElementById("view-map"),
        mapOptions);
    var marker = new google.maps.Marker({
    position: simple,
    map: map,
    title: 'Hello Karachi!'});  
}
google.maps.event.addDomListener(window, 'load', initialize);

can i move php array $row['l_longitude'] and $row['l_latitude'] directly to my javascript? i have google it alot but have not found solution yet. i want to alert that array in my javascript at "alert array here"

解决方案

If you wish to print a complex structure into your page that is worked in a browser, you need a syntax that Javascript can parse, and the serverside language can print.

The neareset possibilities are:

  • XML
  • JSON

I recommend JSON for now.

So with PHP, you write somewhere:

echo "var myTransportedArrayJson = \"" . json_encode($my_array) . "\"";

And then let the Javascript parse your json expression:

var myTransportedArray = JSON.parse(myTransportedArrayJson);
alert(myTransportedArray);   // <<< and here you are.

Of course the snippet must be part of the response, otherwise it won't reach the server. Also, you should embrace the parsing with try .. catch. If you go on and work with even more complex structures, you have to consider escaping certain chars. So you will find now some literature :-)

这篇关于我可以从PHP数组中移动数组从JavaScript数据库回收的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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