Question

我正在尝试创建一个与我的数据库通信的动态AJAX搜索栏。这是我的代码。

function getXmlHttpRequestObject(){
if(window.XMLHttpRequest){
    return new XMLHttpRequest();
}
else if (window.ActiveXObject){
    return new ActiveXObject("Microsoft.XMLHTTP");
}
else{
    alert("Your browser does not support our dynamic search");
}
}

var search = getXmlHttpRequestObject();

function ajaxSearch(){
    if (search.readyState == 4 || search.readyState == 0){
        var str = escape(document.getElementById('searchBox').value);
        search.open("GET", 'searchSuggest.php?search=' + str, true);
        search.onreadystatechange.handleSearchSuggest;
        search.send(null);
    }
}

function handleSearchSuggest(){
    if(search.readyState == 4){
        var ss = document.getElementById('ajaxSearch');
        ss.innerHTML = '';
        var str = search.responseText.split("\n");
        for(i=0; i<str.length-1; i++){
            var suggestion = '<div onmouseover="javascript:suggestOver(this);"';
            suggestion += 'onmouseout="javascript.suggestOut(this);"';
            suggestion += 'onclick="javascript:setSearch(this.innerHTML);"';
            suggestion += 'class="suggestLink">' + str[i] + '<div>';
            ss.innerHTML += suggestion;
        }
    }
}

function suggestOver(divValue){
    divValue.className = "suggestLink";
}

function suggestOut(divValue){
    divValue.className = "suggestLink";
}

function setSearch(x){
    document.getElementById('searchBox').value = x;
    document.getElementById('ajaxSearch').innerHTML = '';
}

问题是readyState从0变为1，但随后它不会变为任何其他状态。我需要将其更改为4以进入函数 handleSearchSuggest（）。我也在控制台中收到此错误： TypeError：search.onreadystatechange为null

Answer 1

您需要正确设置回调功能。

search.onreadystatechange = handleSearchSuggest;

请注意，就绪状态为1表示OPENED，4表示DONE。您可以通过XMLHttpRequest类的属性进行测试：

XMLHttpRequest.UNSENT == 0
XMLHttpRequest.OPENED == 1
XMLHttpRequest.HEADERS_RECEIVED == 2
XMLHttpRequest.LOADING == 3
XMLHttpRequest.DONE == 4

Answer 2

尝试：

search.onreadystatechange = handleSearchSuggest;

readyState不会更改为4

2 个答案: