aldm
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Hi, I have a problem. I'm working on one project. I want to find file which contains site logo words. Ftp has a tone of files, so I can't search file by file. I guess there is no way to get content of .php file in string? So I can set it in loop and compare content of all files with related words. Is there any tool which I can use for this? Thanks in advance, Amer
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yes...for example, I want to change innerHTML of some element in child page, with code from parent page...how can I do that?
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Hi, Is there any way (an how) to access from one html document to another html document with javascript (or php)? For example, I have html document named dodaj_novost.html and another document named video_index.html. If I want to change the innerHTML of element in video_index.html from dodaj_novost.html? Thanks in advance
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Hi, I'm building one application with HTML, CSS and PHP. In this web application, I have a form for creating new member and form with list of members, where member should be selected from the "select" element in html form, and then informations about that member can be viewed. How can I add member's name and surname to "select" when I'm creating him, on form for view member's profile? Thanks in advance
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I tried that, but it shows me the same error again
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When I want to create this procedure: create procedure Ubaci(isbna char(13), autora char(30), naslova char(60), cijenaa float(4,2)) begin insert into knjige(isbn, autor, naslov, cijena) values(isbna, autora, naslova, cijenaa); end in mysql, it showes me error: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 4 What's wrong? Thanks in advance
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[SOLVED] problem with inserting data in mysql database
aldm replied to aldm's topic in PHP Coding Help
Problem is solved by changing line $ubaci="insert into knjige values(".$id.",".$autor.",".$naziv.",".$cij.")"; with: $ubaci="insert into knjige values('$id','$autor','$naziv','$cij')"; Thank for help, anyway -
I'm building one application for book salary in PHP. I have a HTML form with code: <html> <head> <title>Nova knjiga</title> </head> <body> <h1>Unos nove knjge</h1> <br><br> <form action="unos.php" method="POST"> ISBN: <input name="id" type="text"> <br><br> Autor: <input name="autor" type="text"> <br><br> Naziv: <input name="naziv" type="text"> <br><br> Cijena: <input name="cij" type="text"> <br><br><br> <input type="reset" value="Ponovni unos"> <input type="submit" value="Unesi"> </form> </body> Code of unos.php file: <?php $id=$HTTP_POST_VARS["id"]; $autor=$HTTP_POST_VARS["autor"]; $naziv=$HTTP_POST_VARS["naziv"]; $cij=$HTTP_POST_VARS["cij"]; if(!$id || !$autor || !$naziv || !$cij){ echo 'Niste unijeli sve podatke'; echo '<br>Pokusajte ponovo'; exit; } @ $db=mysql_connect('localhost', 'root',''); if(!$db){ echo 'Greska prilikom prikljucenja na bazu'; echo 'Pokusajte ponovo'; exit; } mysql_select_db('knjige_db', $db); $id=addslashes($id); $autor=addslashes($autor); $naziv=addslashes($naziv); $cij=addslashes($cij); $ubaci="insert into knjige values(".$id.",".$autor.",".$naziv.",".$cij.")"; mysql_query($ubaci); mysql_close($db); echo 'Knjiga uspjesno dodana'; ?> When I fill the field and click yes, there is no new record in database. Why???
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it works now, thanks everyone for help
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I save both files (forma.html and phpinfo.php) in F:\xampp\htdocs and action is "phpinfo.php". When I type localhost/phpinfo.php in my browser, it works, but with click on button, it doesn't works. It shows me a blank page with adress: file:///F:/xampp/htdocs/phpinfo.php
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When I change path ("F:\xampp\htdocs\phpinfo.php") to "phpinfo" and click on "Posalji" button, I get blank page. I also changed phpinfo.php file to: <?php echo 'Something'; ?> but with no result (still, there is a blank page when I click on button)
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Hi, I have Apache server installed on my computer and I'm learning PHP. I have a HTML form with following code: <html> <head> <title>Forma</title> </head> <body> <form action="F:\xampp\htdocs\phpinfo.php" method="POST"> <br>Unesite tekst: <br><input type="text" name="podatak" value=""> <br><input type="submit" name="submit" value="Posalji"> </form> </body> </html> Code of phpinfo.php file: <?php phpinfo(); ?> When I click on "Posalji" button, firefox gives me next alert: Firefox doesn't know how to open this address, because the protocol (f) isn't associated with any program. Has anyone idea how could I solve this problem? Thanks in advance