2012-07-05 105 views

回答

0

有一次,我发现自己在同样的情况你。另外,我有很多实际发现的元素,但都不可见 - JSF还不够快,无法让它们可见。另外,我厌倦了一次又一次地写selenium

于是我坐下来,写了下面的代码。它等待所有元素出现在页面上并且在与它们交互之前可见(或者在超时后失败)。因为我已经搬到了webdriver的,所以我没有原代码,但它是这样的:

public static long WAIT = 10000; // ten seconds 

private void waitForElement(String locator) { 
    long targetTime = System.currentTimeMillis() + WAIT; 
    boolean found; 
    do { 
     found = selenium.isElementPresent(locator) && selenium.isVisible(locator); 
    } while (!found && (targetTime < System.currentTimeMillis())); 
    if (!found) { 
     throw new SeleniumException("Element " + locator + " not found"); 
    } 
} 

public void click(String locator) { 
    waitForElement(locator); 
    selenium.click(locator); 
} 

public void type(String locator, String text) { 
    waitForElement(locator); 
    selenium.type(locator, text); 
} 

至于waitForCondition()这应该是一个代码检测一个元素是否存在:

String locator = "id=anything"; 
String script = 
     "var retValue = true;" + 
     "try {" + 
     " selenium.browserbot.findElement('" + locator + "');" + 
     "} catch(e) {" + 
     " retValue = false;" + 
     "}" + 
     "retValue;"; 
selenium.waitForCondition("!!selenium.browserbot.findElement('" + locator + "')", "10000"); 
selenium.click(locator); 

,只是普通的JavaScript:

var retValue = true; 
try { 
    selenium.browserbot.findElement('" + locator + "'); 
} catch(e) { 
    retValue = false; 
} 
retValue; 
1

我调查了生成的html的A4J:状态。下面的代码现在完成这项工作,它比wait()声明更好,但我正在寻找更好的解决方案。

// depends on <a4j:status> present in the page under test 
selenium.waitForCondition(   
    "selenium.browserbot.getCurrentWindow().document.getElementById(
    "_viewRoot:status.start\").style.display == 'none'", 
    "3000");