How to call this function with url parameter
function test($str, $str_){
if ($str == $str)
echo "null";
else
echo "helloworld";
}
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Please define "doesn't work".deceze– deceze ♦2011-08-22 23:54:05 +00:00Commented Aug 22, 2011 at 23:54
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well as said cant send variables to function test($str, $str_), i need to send 2 variables with url link to the function.User6996– User69962011-08-22 23:58:25 +00:00Commented Aug 22, 2011 at 23:58
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I update my reply taking in care this new information. Maybe that helps.leticia– leticia2011-08-23 00:16:33 +00:00Commented Aug 23, 2011 at 0:16
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4 Answers
Use this:
switch($_GET['cmd']) {
case 'hello':
test($_GET['cmd'],'second_parameter_value');
}
The problem is that you aren't passing a value for the second parameter. I execute previous code and prints "helloworld"
Second part:
If your intention is to call the function using two parameters in the Url you can use the follow (I'm only modifying the important parts in your own initial code):
function test($str, $str_){
if ($str == "null")
echo "null";
else
echo "helloworld";
}
switch($_GET['cmd']) {
case 'hello':
test($_GET['cmd'],$_GET['cmd2']);
}
and the Url to call this is: execute.php?cmd=hello&cmd2=hello2
5 Comments
Diosney
Great! What a simple answer :D
leticia
I recommend use what DaveRandom suggest when you are developing: "turn on display_errors in php.ini, or you can put the line ini_set('display_errors',TRUE); at the top of your script" and remove this settings before final users use your scripts, to avoid show them possible important information about your scripts or server.
leticia
Power-Mosfet, answering your comment: your function is declared to receive two parameters, but you are calling the function passing to it only one. Which is your intention with the second parameter called "$str_"?
User6996
Im aware of this. I have modified my question, how to call test($str, $str_) with a url link
leticia
I modify the reply with "Second part" section, that is what you mean?
The function is defined with two parameters, but you have only passed in one.
This will cause a fatal error and PHP stop execution - you should get an error message to this effect, if you are not it would be advisable for you to turn on display_errors in php.ini, or you can put the line ini_set('display_errors',TRUE); at the top of your script.
Try this:
function test ($str) {
if ($str == "null") {
echo "null";
} else {
echo "helloworld";
}
}
switch ($_GET['cmd']) {
case 'hello':
test($_GET['cmd']);
break;
default:
echo "No match in switch structure";
}
Comments
function Init()
{
if(isset($_GET['cmd']))
{
$command = $_GET['cmd'];
switch($command)
{
case 'hello':
test($command);
break;
default:
echo 'hello world';
}
} else {
echo 'Enter Command';
}
}
function test($cmd)
{
echo 'Command: ' . $cmd;
}
Init();
Maybe this can help you
3 Comments
DaveRandom
No explanation of your very modified code for an obvious newb?
Senad Meškin
Does anyone needs explanation for this??
DaveRandom
I don't, but since the OP didn't clock the fact that he wasn't passing a second parameter to the function, he might. No offence meant, but a couple of comments in the code cost nothing... :-)
Try this one
$urlParams = explode('/', $_SERVER['REQUEST_URI']);
$functionName = $urlParams[2];
$functionName($urlParams);
function func1 ($urlParams) {
echo "In func1";
}
function func2 ($urlParams) {
echo "In func2";
echo "<br/>Argument 1 -> ".$urlParams[3];
echo "<br/>Argument 2 -> ".$urlParams[4];
}
and the urls can be as below
http://domain.com/url.php/func1
http://domain.com/url.php/func2/arg1/arg2
