hey
i learned that c programm can be executed using shell script..i have done it and it is working well.but now im trying to do similar thing using shell execution in php.but i got a problem.
this is what im trying to do
for example
i have this c program
#include <stdio.h>
#include <stdlib.h>
int main(int arg,char *argv[])
{
int option=atoi(argv[1]);
int i=0;
int d[10]={1,2,3,4,5,6,7,8,9,10};
FILE *fp;
char file_one[]="text.txt";
char file_second[]="text1.txt";
if(option==1)
{
fp=fopen(file_one,"a");
if(fp==NULL) exit(1);
}
if(option==2)
{
fp=fopen(file_second,"a");
if(fp==NULL) exit(1);
}
for(i=0;i<10;i++){
fprintf(fp,"x[%d]:%d,",i,d[i]);
}
return 0;
}
I can execute this c program and passing the value to it using this shell script:
#!/bin/sh
echo -n "enter the value to pass to c programm : "
read var
echo "The value is : $var"
c=$(./myfile $var)
As a result,this shell script will execute c program and then text.txt will be presented in my directory if var=1 or text1.txt if var=2.
After that i tried to do similar things using php
this is my code:
<html>
<body>
<form method="post">
<p>Num1:<input type="text" name="arg1" size="40" value=""/></p>
<p><input type="submit" name="submit" value="GO"/>
</form>
<?php
if(isset($_POST['arg1']))
{
$num1=$_POST['arg1'];
if($num1){
$output=shell_exec("./myfile $num1");
print "<p>$output</p>";
}
}
?>
</body>
</html>
But it doesnt give me the same result as shell script.why?did i do something wrong in my php code?
is the command shell_exec("./myfile $num1") similar to $(./myfile $my_var)?i assumed that these both commands are similar.am i wrong?hope someone can guide me to make my program work.thanks in advanced.
