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我從zend.com及其示例下載了庫Yotube API PHP(搜索視頻)。 (在Windows本地主機)一切正常Debian的服務器上的微細... 上傳,並有在請求搜索視頻提供了一個錯誤:從視頻browser.js這個錯誤的Youtube API PHP問題
Invalid response received - Status: 404
代碼
/**
* Sends an AJAX request to the server to retrieve a list of videos or
* the video player/metadata. Sends the request to the specified filePath
* on the same host, passing the specified params, and filling the specified
* resultDivName with the resutls upon success.
* @param {String} filePath The path to which the request should be sent
* @param {String} params The URL encoded POST params
* @param {String} resultDivName The name of the DIV used to hold the results
*/
ytvbp.sendRequest = function(filePath, params, resultDivName) {
if (window.XMLHttpRequest) {
var xmlhr = new XMLHttpRequest();
} else {
var xmlhr = new ActiveXObject('MSXML2.XMLHTTP.3.0');
}
xmlhr.open('POST', filePath, true);
xmlhr.setRequestHeader('Content-Type', 'application/x-www-form-urlencoded');
xmlhr.onreadystatechange = function() {
var resultDiv = document.getElementById(resultDivName);
if (xmlhr.readyState == 1) {
resultDiv.innerHTML = '<b>Loading...</b>';
} else if (xmlhr.readyState == 4 && xmlhr.status == 200) {
if (xmlhr.responseText) {
resultDiv.innerHTML = xmlhr.responseText;
}
} else if (xmlhr.readyState == 4) {
alert('Invalid response received - Status: ' + xmlhr.status);
}
}
xmlhr.send(params);
}
的英語不好,對不起
發佈您的代碼。 – 2010-02-18 19:07:35