[英]Having problem in fetching data from database using PHP MySQL PHP
I have tried a very simple example of PHP MySQL Ajax following the example - 我在该示例之后尝试了一个非常简单的PHP MySQL Ajax示例-
http://www.tutorialspoint.com/ajax/ajax_in_action.htm http://www.tutorialspoint.com/ajax/ajax_in_action.htm
I created two files putting these in a same folder and uploaded it to the server.I have also checked the connection with DB from the php file. 我创建了两个文件,并将它们放在同一文件夹中,然后将其上传到服务器。我还从php文件中检查了与DB的连接。 But I don't understand why the html file does not bring any data from the the database.
但是我不明白为什么html文件不带数据库中的任何数据。 What's wrong with that.
怎么了 The example i followed can be found here.
我遵循的示例可以在这里找到。
http://aiworker2.usask.ca/age/ajax.html http://aiworker2.usask.ca/age/ajax-example.php http://aiworker2.usask.ca/age/ajax.html http://aiworker2.usask.ca/age/ajax-example.php
I'm very new in AJAX and really don't understand what's going on. 我是AJAX的新手,真的不了解发生了什么。 Any sort of help will be appreciated.
任何帮助将不胜感激。
My 2 cents: First of all, try a simpler Ajax request to get yourself used to the mechanics of the thing. 我的2分钱:首先,尝试一个更简单的Ajax请求,以使自己熟悉事物的原理。
All you have to do on the PHP side is echo
out some data. 在PHP方面,您要做的就是
echo
一些数据。 For example, you could have your ajax open "test.php", then have "test.php" look like this: 例如,您可以将ajax打开“ test.php”,然后将“ test.php”显示如下:
<?php echo "working"; ?>
Then, just alert yourRequest.responseText
(the response text property) and it should say "working". 然后,只需警告
yourRequest.responseText
(响应文本属性),它应该显示“正在工作”。
Once you've got this moving smoothly, you can modify the PHP file to make a MySQL request. 一旦顺利进行,就可以修改PHP文件以发出MySQL请求。
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