Paldo Posted May 19, 2009 Share Posted May 19, 2009 Hi I would like to ask how to select only one thing from database co I can use it later on. I have a databaze containing name, login, password, account number, balance. What I need is get only account number and store it to variable, for example $balance. How can I do this. Thanks Quote Link to comment Share on other sites More sharing options...
JD* Posted May 19, 2009 Share Posted May 19, 2009 $result = mysql_query("SELECT account_number FROM database") or die(mysql_error()); $balance = mysql_result($result, 0, "account_numer"); echo $balance Quote Link to comment Share on other sites More sharing options...
Paldo Posted May 19, 2009 Author Share Posted May 19, 2009 Well tahnks I did this: $result = mysql_query("SELECT account_number FROM database") or die(mysql_error()); $balance = mysql_result($result, 0, "account_numer"); echo $balance and it's working. BUT one last thing I would like to ask. What is a correct way to send value of this $balance to another page so I can use it there???? I tried : echo'<a href="http://parobek.euweb.cz/transfer.php?t=$balance">Prevody</a>'; but when I echo t on second page it prints $balance not the value of it. Thanks Quote Link to comment Share on other sites More sharing options...
JD* Posted May 19, 2009 Share Posted May 19, 2009 If you pass it through the url, you need to do this: $balance = $_GET['t']'; echo $balance Quote Link to comment Share on other sites More sharing options...
Paldo Posted May 19, 2009 Author Share Posted May 19, 2009 Hey! Shame on me that I'm putting this question on you, but I just cant solve it and I'm realy desprete by now... it says: Parse error: parse error, expecting `','' or `';'' in /3w/euweb.cz/p/parobek/welcome.php on line 31 <html> <body> <?php if ($name=="admin" && $password=="adnddnd" ) { print "welcome Admin /n"; echo'<a href="http://parobek.euweb.cz/new.php">Creating new customer</a>'; echo'<a href="http://parobek.euweb.cz/zmena.php">Information update</a>'; echo'<a href="http://parobek.euweb.cz/odstran.php">to delete a customer</a>'; } else { $con = mysql_connect("mysql.webzdarma.cz","parobek","adnddnd"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("parobek", $con); $result = mysql_query("SELECT * FROM Customers WHERE Login='$meno' AND pasword='$heslo'"); while($row = mysql_fetch_array($result)) { echo $row['Name'] . " " . $row['Login']. " " . $row['Password']. " " . $row['accountID']. " " . $row['balance']; echo "<br />"; $resul = mysql_query("SELECT stav FROM Customers WHERE Login=$name ") or die(mysql_error()) ; $balance = mysql_result($resul, 0, "balance"); echo'<a href="http://parobek.euweb.cz/transfer.php?t='$balance'">Transfers</a>'; } echo $balance; mysql_close($con); } ?> </body> </html> Quote Link to comment Share on other sites More sharing options...
JD* Posted May 19, 2009 Share Posted May 19, 2009 This line: echo'<a href="http://parobek.euweb.cz/transfer.php?t='$balance'">Transfers</a>'; should be echo'<a href="http://parobek.euweb.cz/transfer.php?t=$balance">Transfers</a>'; you don't need quotes around url variables Quote Link to comment Share on other sites More sharing options...
Paldo Posted May 19, 2009 Author Share Posted May 19, 2009 Well thanks that worked well. I moved a bit further... Yet something always not working. As I try to finaly use the transfered variable on a target site it says: Parse error: parse error in /3w/euweb.cz/p/parobek/transfer.php on line 16 <html> <head> </head> <body> <font size="12" color="red" face="Arial, Helvetica">Zadajte cislo uctu kam previest zdroje</font> <form action="http://www.parobek.euweb.cz/transferconfirm.php" method="post"> Cislo uctu: <input type="int" name="cisuctu" /><br /> Ciastka k prevodu: <input type="int" name="prevod" /> <input type="submit" /> </form> <?php $balance = $_GET['t']'; echo $balance; ?> </body> </html> I realy don't know whats wrong with it... If you have a minute to ceck it out I would be thankfull... Quote Link to comment Share on other sites More sharing options...
JD* Posted May 19, 2009 Share Posted May 19, 2009 You have an extra ' at the end of your balance variable Quote Link to comment Share on other sites More sharing options...
Paldo Posted May 19, 2009 Author Share Posted May 19, 2009 Great! Thanks! But why its printig $balance instead of value a was trying to transfer from first page? Quote Link to comment Share on other sites More sharing options...
JD* Posted May 19, 2009 Share Posted May 19, 2009 Sorry, I didn't see in your original code that you were echoing your link with single quotes. Change this: echo'<a href="http://parobek.euweb.cz/transfer.php?t='$balance'">Transfers</a>'; to this: echo'<a href="http://parobek.euweb.cz/transfer.php?t='.$balance.'">Transfers</a>'; Quote Link to comment Share on other sites More sharing options...
Paldo Posted May 19, 2009 Author Share Posted May 19, 2009 Wow man you are some kind of genius or what ?!? Thanks a lot. A LOT! Quote Link to comment Share on other sites More sharing options...
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