2014-12-31 38 views
-2

嗨我是JSP新手,我試圖將W3School的Ajax示例轉換爲jsp,但我無法這樣做。任何人都可以幫忙。將php/asp ajax示例轉換爲jsp

這裏是PHP例子

http://www.w3schools.com/php/php_ajax_php.asp

這裏是ASP例子

http://www.w3schools.com/ajax/ajax_asp.asp

我試圖用java的子功能,但未能如願。

在此先感謝。

+0

這些例子可以在JSP中直接運行,如果有'xmlhttp.open(...)'方法。你試過了嗎? – Tiny

+0

我正在尋找下面的代碼在jsp <?php $ q = $ _REQUEST [「q」]; $ hint =「」; if($ q!==「」){ $ q = strtolower($ q); $ len = strlen($ q); ($ h as == $「name){ $ hint = $ name;} {其他{ }其他{hint。=「,$ name」; } } } } echo $ hint ===「」? 「沒有建議」:$ hint; ?> –

回答

0

我的理解是,你想要將PHP/ASP代碼轉換爲jsp。這裏是jsp代碼。請保存此代碼gethint.jsp頁面,並在HTML頁面更改URL僅通過改變URL和查詢字符串參數(S)爲目標的jsp頁面

<%@page import="java.util.Arrays"%> 
<% 
// Array with names 
String hint = " "; 
String[] a = { 
    "Anna","Brittany","Cinderella","Diana","Eva","Fiona","Gunda", 
    "Hege","Inga","Johanna","Kitty","Linda","Nina","Ophelia" 
    ,"Petunia","Amanda","Raquel","Cindy","Doris","Eve","Evita","Sunniva" 
    ,"Tove","Unni","Violet","Liza","Elizabeth","Ellen","Wenche","Vicky" 
    }; 
    // get the q parameter from URL 
    String q = request.getParameter("q"); 

    if (q != " ") { 
    for(String name : a) 
    { 
    if(name.toLowerCase().startsWith(q.toLowerCase())) 
    { 
     if(hint == " ") 
     { 
      hint = name; 
     } 
     else 
     { 
      hint = hint + ", " + name; 
     } 
    } 
    } 
} 

if(hint == " ") 
{ 
    out.println("No suggestion");  
} 
else{ 
out.println(hint); 
    } 

    %> 
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謝謝Dulton。代碼對我來說工作正常。非常感謝 –