Question

以下是我已经使用的脚本，它只是将内容放在uint8_t globalColorTableFlag = 1; uint8_t colorResolution = 001; uint8_t sortFlag = 0; uint8_t sizeOfGlobalColorTable = 001; uint32_t packed = ((globalColorTableFlag << 24) | (colorResolution << 16) | (sortFlag << 8) | (sizeOfGlobalColorTable << 0)); NSLog(@"%d",packed); // Logs 16842753, should be: 10010001 NSLog(@"0x%02X",packed); // Logs 0x1010001, should be: 0x91

中

的index.php

response.php

response.php

<input type="button" id="fire" value="Show Items">
<script src="http://code.jquery.com/jquery-1.7.2.js"></script>
<script type="text/javascript">
    $("#fire").click(function(){
        $.ajax({
            type: "POST",
            url : "response.php",
             data: { 
            'test': 'test', 
        },
            success : function(data)
            {

            $("#result").html(data);
            }
        },"json");
    });
    </script>

<div id="result"></div>

我尝试通过此代码制作google autosuggest

<?php
echo '

    <input type="text" id="startingloc">
    <input type="text" id="endingloc">';
?>

我正在

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.2/jquery.min.js"></script>
<script type="text/javascript" src="http://maps.googleapis.com/maps/api/js?sensor=false&libraries=places"></script>

    <script type="text/javascript">
    $("#startingloc").on("input",function() {

        //alert("Change detected!");
        console.log("Changed");

    });
        google.maps.event.addDomListener(window, "load", function () {
            var places = new google.maps.places.Autocomplete(document.getElementById("startingloc"));
                  var places = new google.maps.places.Autocomplete(document.getElementById("endingloc"));
            google.maps.event.addListener(places, "place_changed", function () {
                var place = places.getPlace();
                var address = place.formatted_address;
                var latitude = place.geometry.location.k;
                var longitude = place.geometry.location.D;
                var mesg = "Address: " + address;
                mesg += "\nLatitude: " + latitude;
                mesg += "\nLongitude: " + longitude;
            });
        });

    </script>
    <input type="text" id="startingloc">
    <input type="text" id="endingloc">

当我尝试在主页中添加javascript时，我没有在那里获得自动提示。

我该如何解决这个问题？

以下是我的最新代码

的index.php

Failed to execute 'write' on 'Document': It isn't possible to write into a document from an asynchronously-loaded external script unless it is explicitly opened.

response.php

<input type="button" id="fire" value="Show Items">
<script src="http://code.jquery.com/jquery-1.7.2.js"></script>
<script type="text/javascript" src="http://maps.googleapis.com/maps/api/js?sensor=false&libraries=places"></script>
<script type="text/javascript">
    $("#fire").click(function(){
        $.ajax({
            type: "POST",
            url : "response.php",
             data: { 
            'test': 'test', 
        },
            success : function(data)
            {

            $("#result").html(data);
            }
        },"json");
    });
    </script>
<div id="result"></div>

Answer 1

只需删除

即可

   google.maps.event.addDomListener(window, "load", function () {

[并且结束});]

在response.php

中

并且不要尝试从您的ajax页面加载.js并始终从父页面加载它。

所以你的代码将是

response.php

<?php
echo '
<script type="text/javascript">
            var places = new google.maps.places.Autocomplete(document.getElementById("startingloc"));
            var places = new google.maps.places.Autocomplete(document.getElementById("endingloc"));
            google.maps.event.addListener(places, "place_changed", function () {
                var place = places.getPlace();
                var address = place.formatted_address;
                var latitude = place.geometry.location.k;
                var longitude = place.geometry.location.D;
                var mesg = "Address: " + address;
                mesg += "\nLatitude: " + latitude;
                mesg += "\nLongitude: " + longitude;
            });
    </script>
    <input type="text" id="startingloc">
    <input type="text" id="endingloc">';
?>

您的index.php将保持不变

将动态内容加载到页面

1 个答案: