IamTomCat Posted January 24, 2015 Share Posted January 24, 2015 Hi I want to echo out something when I post the value of a radio button and this value matches the same name of a html file. How can I formulate my code to achieve that? What I have so far: <?php if (isset($_POST['submitradio'])) { $selected_dir = $_POST['radio1']; echo substr($selected_dir, 0, -4); echo '<img src="'.$selected_dir.'" />'; } ?> So far I get for example the name of the image(order.gif) and the image order. gif. What I now want is to formulate my code so that I can say if I have an image called "order.gif" and a file called "order.html" that I can then echo out some thing. I don't want to actually name the image or the html file. I just want to say if I have two different file types that start with the same name then I can echo out what ever. How can I do that? Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/ Share on other sites More sharing options...
Ch0cu3r Posted January 24, 2015 Share Posted January 24, 2015 (edited) Not sure what you are asking but maybe... use file_exists to see if a html file shares the same name as the image. if (isset($_POST['submitradio'])) { $selected_dir = $_POST['radio1']; echo '<img src="'.$selected_dir.'" />'; // get the filename of the file $filename = pathinfo($selected_dir, PATHINFO_FILENAME); // check to see if a html file named the same also exists if(file_exists("$filename.html")) { echo "$filename.html shares the same name as $selected_dir"; } else { echo "Could not find a html file that shares the same $filename"; } } Edited January 24, 2015 by Ch0cu3r Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/#findComment-1504089 Share on other sites More sharing options...
IamTomCat Posted January 24, 2015 Author Share Posted January 24, 2015 HI many thanks for your reply. I try to explain it again. I want that when I submit the radio button($selected_dir = $_POST['radio1']), ( which would be for example order.gif) and when there is a file in the same folder called "order.html" then I could for example echo out "boa boa". If I submit let's say a radio button value (let's say "smoke.gif") and there is no html file called "smoke.html" then I could echo out "No boa boa available". Don't know how else to explain it. It's just really important that I don't give actually the name of each file. I want that the programme looks what is there. Example of what I got with the code you gave me: Could not find a html file that shares the same order. But there is a file called order.gif and a file called order.html in the same folder. Help would be very much appreciated. Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/#findComment-1504091 Share on other sites More sharing options...
Ch0cu3r Posted January 24, 2015 Share Posted January 24, 2015 Thats is what my code does. All you need to do is change the echo statements to what you want to output when a filename match is/not found. For my code to work correctly the .gif and .html files need to be in the same directory as where this code is being ran. If the files are in a different directory then you need to tell us. Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/#findComment-1504093 Share on other sites More sharing options...
IamTomCat Posted January 24, 2015 Author Share Posted January 24, 2015 Hi. Thanks for your reply and your patience. Well the files are in the same folder. And I don't get anything for that part: echo "$filename.html shares the same name as $selected_dir"; Have you tried it yourself? And when yes, did it work for you? Well I just wonder, if I may have made a mistake or if there could be maybe an other solution for the problem. I tried now for hours without making any progress. Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/#findComment-1504097 Share on other sites More sharing options...
Ch0cu3r Posted January 24, 2015 Share Posted January 24, 2015 Yes. I have tested my code and it functions correctly. This is the form I am using to test with and I only have an order.html file in the same directory as the php code <form method="post"> <input type="radio" name="radio1" value="order.gif" /> Order<br /> <input type="radio" name="radio1" value="dummy.gif" /> Dummy<br /> <input type="submit" name="submitradio" /> </form> Selecting the Order radio button I get "order.html shares the same name as order.gif" Selecting Dummy radio button I get "Could not find a html file that shares the same dummy". Maybe the data in your form is different. What is the output of the following printf('<pre>%s</pre>', print_r($_POST, 1)); Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/#findComment-1504098 Share on other sites More sharing options...
IamTomCat Posted January 24, 2015 Author Share Posted January 24, 2015 The output of : printf('<pre>%s</pre>', print_r($_POST, 1)); is: Array ([radio1] => Animation/Despicable Me.jpg[submitradio] => submit) Animation is the folder name. Despicable Me.jpg is the image name. Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/#findComment-1504100 Share on other sites More sharing options...
Ch0cu3r Posted January 24, 2015 Share Posted January 24, 2015 Animation is the folder name That will be why my code is not working as expected. You did not mention that earlier As I said earlier all files (including the php file) need to all in the same directory in order for it work correctly. Code updated if (isset($_POST['submitradio'])) { $selected_file = $_POST['radio1']; // get the filename of the file $fileinfo = pathinfo($selected_file); $filename = $fileinfo['dirname'] . DIRECTORY_SEPARATOR . $fileinfo['filename']; // check to see if a html file named the same also exists if(file_exists("$filename.html")) { echo "$filename.html shares the same name as $selected_file"; } else { echo "Could not find a html file that shares the same name as $filename.html"; } } Should work as expected this time. Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/#findComment-1504102 Share on other sites More sharing options...
IamTomCat Posted January 24, 2015 Author Share Posted January 24, 2015 It worked . Thank you so much Quote Link to comment https://forums.phpfreaks.com/topic/294202-echo-out-something-when-selected_dir-matches-the-same-name-first-part-of-a-html-file/#findComment-1504103 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.