将PHP变量用于JS confirm()方法

时间:2016-04-07 07:45:55

标签: javascript php variables confirm

这是一个简单的问题,我有一个变量(一个字符串),根据用户操作在上一页的内容而变化,它可以是' package',' content&#39 ;等

由于修改可能会产生影响,我想通过使用确认方法确保用户知道他的行为,确认方法,"您确定要删除此包"如果这是一个包或者"你是......这个内容"等

在我的JS中我做到了:

#url = raw_input('Enter - ')
url = 'http://www.columbia.edu/kermit/k95.html'
itr = raw_input('Enter iteration: ')
i = int(itr)

n = raw_input('Enter Number: ')
n = int(n)

html = urllib.urlopen(url).read()
soup = BeautifulSoup(html)
tags = soup('a')

print 'Link:' , url
while i > 0:
    i = i - 1
    if i == 0:
        break
    for tag in tags:  
        me = tag.get('href', None)
        #Just to make sure the link/content match print tag.contents[0]
        link = tags[(n - 1)]
        #print link 
    links = link.get('href', None)
    print 'Link:', links

Enter - http://www.columbia.edu/~fdc/
Enter count: 4
Enter Position: 9
Link: http://www.columbia.edu/~fdc/
Link: http://www.columbia.edu/kermit/k95.html
Link: http://www.columbia.edu/kermit/k95.html (Should be k95faq.html)
Link: http://www.columbia.edu/kermit/k95.html (Should be ckfaq.html)

我想这样做:

var sCategorieType = $(this).closest('tr').attr('name');

这不起作用,我想知道为什么并有另一种选择。

2 个答案:

答案 0 :(得分:1)

javascript中的连接与PHP中的连接不同。在javascript中,您可以将字符串与+符号连接起来。

尝试:

var bConfirm = confirm("Are you sure you want to delete this " + sCategorieType + "? ");

答案 1 :(得分:1)

将其与“+”连接,而不是与“。”连接。符号,它会好的! :)

var bConfirm = confirm("Are you sure you want to delete this " + sCategorieType + "? ");

祝你好运!)