[sampkinq]

Solved

[SyS]

of course you can example code

PHP код:

new bool:flag=false;
for(new i=0;i<sizeof inputtext;i++)
{
if(inputtext[i]=='@')
{
flag=true;
break;
}
}
if(!flag)
{
//if the inputted string does not have @ do the correspondense
} 


[sampkinq]

Quote:

Originally Posted by Sreyas of course you can example code PHP код: new bool:flag=false; for(new i=0;i<sizeof inputtext;i++) { if(inputtext[i]=='@') { flag=true; break; } } if(!flag) { //if the inputted string does not have @ do the correspondense }

Код:

warning 224: indeterminate array size in "sizeof" expression (symbol "")

[SyS]

Quote:

Originally Posted by sampkinq Код: warning 224: indeterminate array size in "sizeof" expression (symbol "")

try it

PHP код:

new bool:flag=false; 
for(new i=0;i<=strlen(inputtext);i++) 
{ 
if(inputtext[i]=='@') 
{ 
flag=true; 
break; 
} 
} 
if(!flag) 
{ 
//if the inputted string does not have @ do the correspondense 
} 


NOTE:above code is just example..and it will only detect @ you can think logically and make it work for '.' and other things you want

[sampkinq]

Quote:

Originally Posted by Sreyas try it PHP код: new bool:flag=false;  for(new i=0;i<=strlen(inputtext);i++)  {  if(inputtext[i]=='@')  {  flag=true;  break;  }  }  if(!flag)  {  //if the inputted string does not have @ do the correspondense  }  NOTE:above code is just example..

Thank you for your help.

[SyS]

Quote:

Originally Posted by sampkinq Thank you for your help.

post the main topic back dont remove it so that someone in future can refer here

[Vince]

We should start keeping a list of people who do that, as a reminder to everyone to never reply to their topics again. Deleting your thread after you've got your solution is both egocentric and ungrateful.

[d1git]

Why would you create both an unnecessary loop and variable? Might aswell do this;

Код:

if(strfind(inputtext, "@", true) != -1) {
    // Yes @
} else {
    // No @
}