doogles
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[SOLVED] Setting MySQL table values to PHP variables
doogles replied to doogles's topic in PHP Coding Help
I just added an auto_increment column so I can use a WHERE clause now Thanks for the help. It works now -
[SOLVED] Setting MySQL table values to PHP variables
doogles replied to doogles's topic in PHP Coding Help
hmm...well, I tried that and then echo $name, and nothing showed -
it seems to me you have two else clauses in a row else { echo'<table class="verifytb" style="width:500px"><tr class="tbhd"><td>Emperor Election</td></tr>'; echo'<tr><td class="admincellcen1">You can vote for a capable kingdom <b>in your empire</b> to represent as the Emperor for '.$newarray[empire].'</td></tr>'; // calculate how many ppl vote for this user $check_mine=mysql_query("SELECT COUNT(id) FROM users WHERE Svote= '$newarray[Knumber]'" ,$link ) or die(mysql_error()); $vote_for_me=mysql_fetch_array($check_mine); echo'<tr><td class="admincellleft1">--<b class="highlight">'.$vote_for_me[0].'</b> Kingdoms have voted you for the Emperor.<br>'; // Display current vote for creator if ($newarray['Svote']==0) echo'<span class="txtleft">-You have not vote for the Emperor yet.</span><br>'; else { I just saw your error if ($newarray['Svote']==0) echo'<span class="txtleft">-You have not vote for the Emperor yet.</span><br>'; else { hehe... You're missing the curly bracets on that if clause. And you have a typo! GASP if ($newarray['Svote']==0) {echo'<span class="txtleft">-You have not voted for the Emperor yet.</span><br>';} else {
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you can't have two else clauses in a row it has to be if ($x=1) {do stuff} elseif ($x=2) {do something else} THEN else {do other stuff} You can have more than one elseif, though. If your if clauses aren't related to the same thing, just make them all if clauses if ($x=1) {do that} if ($y=15) {do this} if ($z=45) {do something else}
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Pretty simple question. I think this is more related to PHP than mysql so I posted it here. I just want to select values from mysql and be able to save them to my own PHP variables. The catch is that I don't know what the values will be because they are based on user input, so I can't use a WHERE clause. Any help would be great. The table's name is the only thing known about it. Also, I know there are 3 columns: 'title', 'points_earned', and 'points_possible' and that the table is not empty. All of that is verified. I just need to save the values that are in it to its respective values. Any help is appreciated. I was thinking a few if clauses would work... if(mysql_num_rows(mysql_query(SELECT * FROM $class)) = 1) { //I don't know what to do to save the values to php variables } maybe I need an auto_increment integer to work with?
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you could go into your MySQL DOS box and simply type in your password then enter the following: use [database name]; then do what you wish. Or if you working with PHP, do something like: $conn=mysql_connect("localhost", "[username]", "[password]"); mysql_select_db("[database name]", $conn);
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printing values in a web browser from values in a mysql table
doogles replied to doogles's topic in MySQL Help
let me simplify the question: How do I take values in a mysql table and convert them to readable, changeable PHP variables? -
MySQL version 5.0.22 I want to take a table in mysql and virtually print it in the web browser. BUT I also want to add columns that manipulate the values already in the MySQL table. The user (me) enters their desired table name to be displayed in a small form... <html><head><title>View gradelogs</title></head><body> <h1>View grades</h1> <?PHP $class=$_POST['class']; if ($class =="") { echo "<form method=post> <p>Type the name of the class you want to view:</p><input type=text name=class><input type=submit value=view></form>"; } else { //This is where I want to fetch the MySQL table and display it.} If it helps, here is the table that is created and the script that inserts it (on another file and another form) : $createtable="CREATE table $class ( title varchar (25), points_earned varchar (4), points_possible varchar (4))"; mysql_query($createtable); if ($grade1 !=="") {$ins="insert into $class values('$grade1', '$g1e', '$g1t')"; mysql_query($ins) or die(mysql_error()); //and etc. for $grade2....$grade3..} It's not really a problem. I just want to know how to do it. It'd be easy if it was just one row, but its a whole table. Also, if you could tell me how to save each value in the table to a specific variable, that'd be easier. I can make the table, its just getting the table's values from mysql into a workable variable. Thanks.
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alright, thanks
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so, $test_array=array("$test1","$test2","$test3","$test4"); foreach($test_array as $var) {echo $var;}
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It was hard to explain with just the title, so I figured I'll explain what I mean here: Say I have various variables with the same name and different endings, like $test1, $test2, $test3, $test4, etc... Would it be possible to create a loop that will do something to each of those variables instead of me typing it for each and every one? Something like: for($x=1; $x <=4; $x++) {echo $test$x;} so it would print each test variable, 1-4? Obviously combining the two variables is impossible. Do I need to encase the $x in something like $test".$x." ? Thanks !
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MySQL error in PHP..."Unknown column 'test' in 'fieldlist'"
doogles replied to doogles's topic in MySQL Help
oh, its just on my localhost. I'm not putting this on my site. And thanks! I'll see what that does. -
MySQL version 5.0.22 My problem is when I try to insert values based on user input into a table also based on user input. The table is created fine. I verified it. It's the values I try to put it the table that don't go. Here's the small script that adds the table: $createtable="CREATE table $class ( title varchar (25), points_earned varchar (4), points_possible varchar (4))"; mysql_query($createtable); Here's the script that adds the values: if ($grade1 !=="") {$ins="insert into $class values($grade1, $g1e, $g1t)"; mysql_query($ins) or die(mysql_error());} The error mysql gives depends on the user's value for $grade1. I've verified also that all the values I try to insert exsist by having the script print them to the browser. If $grade1 is 'test' then I get: "Unknown column 'test' in 'fieldlist'" Thanks!