我在我的应用程序中实现谷歌的即时搜索。我希望在用户输入文本时触发HTTP请求。我遇到的唯一问题是，当用户在名字和姓氏之间找到空格时，空格没有被编码为+，从而破坏了搜索。我如何既可以替换空间与一个+，或只是安全URL编码字符串?
$("#search").keypress(function(){
var query = "{% url accounts.views.instasearch %}?q=" + $('#tags').val();
var options = {};
$("#results").html(ajax_load).load(query);
});
当前回答
I'm using MVC3/EntityFramework as back-end, the front-end consumes all of my project controllers via jquery, posting directly (using $.post) doesnt requires the data encription, when you pass params directly other than URL hardcoded. I already tested several chars i even sent an URL(this one http://www.ihackforfun.eu/index.php?title=update-on-url-crazy&more=1&c=1&tb=1&pb=1) as a parameter and had no issue at all even though encodeURIComponent works great when you pass all data in within the URL (hardcoded)
硬编码URL，即>
var encodedName = encodeURIComponent(name);
var url = "ControllerName/ActionName/" + encodedName + "/" + keyword + "/" + description + "/" + linkUrl + "/" + includeMetrics + "/" + typeTask + "/" + project + "/" + userCreated + "/" + userModified + "/" + status + "/" + parent;; // + name + "/" + keyword + "/" + description + "/" + linkUrl + "/" + includeMetrics + "/" + typeTask + "/" + project + "/" + userCreated + "/" + userModified + "/" + status + "/" + parent;
否则不要使用encodeURIComponent，而是尝试在ajax post方法中传递参数
var url = "ControllerName/ActionName/";
$.post(url,
{ name: nameVal, fkKeyword: keyword, description: descriptionVal, linkUrl: linkUrlVal, includeMetrics: includeMetricsVal, FKTypeTask: typeTask, FKProject: project, FKUserCreated: userCreated, FKUserModified: userModified, FKStatus: status, FKParent: parent },
function (data) {.......});
其他回答
更好的办法:
encodeURIComponent转义除以下字符外的所有字符:字母、十进制数字、- _。！ ~ * ' ()
To avoid unexpected requests to the server, you should call encodeURIComponent on any user-entered parameters that will be passed as part of a URI. For example, a user could type "Thyme &time=again" for a variable comment. Not using encodeURIComponent on this variable will give comment=Thyme%20&time=again. Note that the ampersand and the equal sign mark a new key and value pair. So instead of having a POST comment key equal to "Thyme &time=again", you have two POST keys, one equal to "Thyme " and another (time) equal to again.
对于application/x-www-form-urlencoded (POST)，根据http://www.w3.org/TR/html401/interac...m-content-type，空格将被“+”替换，因此可能希望在encodeURIComponent替换之后，将“%20”替换为“+”。
如果您希望更严格地遵守RFC 3986(保留!，'，(，)和*)，即使这些字符没有正式的URI分隔用途，也可以安全地使用以下字符:
function fixedEncodeURIComponent (str) {
return encodeURIComponent(str).replace(/[!'()]/g, escape).replace(/\*/g, "%2A");
}
使用jQuery.param()…… 描述:创建数组、普通对象或适合在URL查询字符串或Ajax请求中使用的jQuery对象的序列化表示。在传递jQuery对象时，它应该包含带有名称/值属性的输入元素。
encodeURIComponent试试。
通过用表示字符的UTF-8编码的一个、两个、三个或四个转义序列替换某些字符的每个实例来对统一资源标识符(URI)组件进行编码(对于由两个“代理”字符组成的字符，只有四个转义序列)。
例子:
var encoded = encodeURIComponent(str);
encodeURIComponent对我来说很好。我们可以在ajax call中给出这样的url。代码如下所示:
$.ajax({
cache: false,
type: "POST",
url: "http://atandra.mivamerchantdev.com//mm5/json.mvc?Store_Code=ATA&Function=Module&Module_Code=thub_connector&Module_Function=THUB_Request",
data: "strChannelName=" + $('#txtupdstorename').val() + "&ServiceUrl=" + encodeURIComponent($('#txtupdserviceurl').val()),
dataType: "HTML",
success: function (data) {
},
error: function (xhr, ajaxOptions, thrownError) {
}
});
试试这个
var query = "{% url accounts.views.instasearch %}?q=" + $('#tags').val().replace(/ /g, '+');
