pjotterke Posted February 26, 2007 Posted February 26, 2007 Hi guys i want to make a prints screen of a screen this is my code... DllCall("captdll.dll", "int", "CaptureScreen", "str", $input_1, "int", 85) But i want to make "$input_1" like this "$input_1.JPG" understand? how can i do that
MHz Posted February 26, 2007 Posted February 26, 2007 If $input_1 is for the name of the picture, then this may do. $input_1 = "PictureName.jpg" DllCall("captdll.dll", "int", "CaptureScreen", "str", $input_1, "int", 85)
pjotterke Posted February 26, 2007 Author Posted February 26, 2007 thanks for the Reply MHz but there is a little prob $input_1 is a variable and iam using it in an other script... so if i use your script he gives a "0" :S:S verry odd... so he wants to use input_1 to times...
Moderators SmOke_N Posted February 26, 2007 Moderators Posted February 26, 2007 thanks for the Reply MHzbut there is a little prob $input_1 is a variable and iam using it in an other script...so if i use your script he gives a "0" :S:Sverry odd...so he wants to use input_1 to times...Look at DLLCall in the help file to understand the return values (You'll also note, that if successful that DLLCall() will return an array). Common sense plays a role in the basics of understanding AutoIt... If you're lacking in that, do us all a favor, and step away from the computer.
pjotterke Posted February 26, 2007 Author Posted February 26, 2007 Look at DLLCall in the help file to understand the return values (You'll also note, that if successful that DLLCall() will return an array).i have looked @ the helpfile...but didn't understand thats why i have asked on the forum i just don't understand how i have to make a jpg with a variable...i can do it if i have a name... like picturethen it's onlue picture.jpg thats easy but with a variable it is different...
Moderators SmOke_N Posted February 26, 2007 Moderators Posted February 26, 2007 i have looked @ the helpfile...but didn't understand thats why i have asked on the forum No kidding... and I gave you an answer directly, but you also need read it... if you can't understand something in the help file, then ask the question specifically on the help file.i just don't understand how i have to make a jpg with a variable...i can do it if i have a name... like picturethen it's onlue picture.jpg thats easy but with a variable it is different...You can't store the picture like you want the way you want in a variable and just display it. There, saved you alot of time.P.S.PaulIA has some screen capture options in his AU3 library in the example scripts forum. Common sense plays a role in the basics of understanding AutoIt... If you're lacking in that, do us all a favor, and step away from the computer.
Shevilie Posted February 26, 2007 Posted February 26, 2007 $input_2 = "PictureName.jpg" DllCall("captdll.dll", "int", "CaptureScreen", "str", $input_2, "int", 85) Start here if you are new Valuater's AutoIT 1-2-3Looking for an UDF - Look hereDo you need to do it twice - Autoit
tAKTelapis Posted February 26, 2007 Posted February 26, 2007 im thinking he has $Input_1 as a user input.. in which case, he needs to "stitch" the .jpg onto the end of it during the Dll call $Input_1 = String("Picture") $input_2 = String($Input_1&".jpg") DllCall("captdll.dll", "int", "CaptureScreen", "str", $input_2, "int", 85)
Shevilie Posted February 26, 2007 Posted February 26, 2007 $pic = GuiCtrlRead($input_1) $pic = $pic & ".jpg" DllCall("captdll.dll", "int", "CaptureScreen", "str", $pic, "int", 85) Start here if you are new Valuater's AutoIT 1-2-3Looking for an UDF - Look hereDo you need to do it twice - Autoit
pjotterke Posted February 26, 2007 Author Posted February 26, 2007 $pic = GuiCtrlRead($input_1)$pic = $pic & ".jpg"DllCall("captdll.dll", "int", "CaptureScreen", "str", $pic, "int", 85)thanks for the reply it works..!!only 1 thing not:(i want to give the JPG a name that i have put in on input_1now it gives a "0" (not good)is that because he thinks he have to give the JPG 2 names ore something elsE?
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