php 通过 URL 传递参数：

Question

mimo.php is my main file... mainjob.php contains the div's that I want to $get

this is an example of my link...

.../pes/mimo.php?rack=1&jobs=0&hosts=1 this means that only entries related to rack 1 will be displayed. The jobs list will not be displayed, but the host menu will be displayed

I'm passing the variables like this...

 echo $_GET["rack"] ;

 echo $_GET["jobs"] ;

 echo  $_GET["hosts"] ;

Rack would echo a grouping of divs with the contents: r1, r2, r3 Jobs it's grouping of divs: j1, j2, j3 Hosts: h1,h2, h3

I'm looking to connect the dots between passing the parameter and getting the data... Anyone feel me?

Best, William

Answer 1

Incorrect: .../pes/mimo.php?rack=1+jobs=0+hosts=1

Correct: .../pes/mimo.php?rack=1&jobs=0&hosts=1

echo $_GET["rack"]; // = 1
echo $_GET["jobs"]; // = 0
echo $_GET["hosts"]; // = 1

Using + is a space operator, meaning that rack = 1 jobs=0 (replace the + with an &)

For the browser, it uses the + for spaces, like &jobs=my+job , and then using $_GET["jobs"] would return my job