I have a php webpage that I want it to show all forum threads at once. He is the current db code i have, it only shows on thread.
mysql_connect("localhost", "", "") or die("could not connect to mysql");
mysql_select_db("") or die("could not connect to db");
$result = mysql_query("SELECT * FROM reference")
or die(mysql_error());
$row = mysql_fetch_array( $result );
echo "<strong>ref_thread_id:</strong> ".$row['ref_thread_id'].'<br />';
echo "<strong>ref_thread_prefix:</strong> ".$row['ref_thread_prefix'].'<br />';
echo "<strong>ref_thread_topic:</strong> ".$row['ref_thread_topic'].'<br />';
echo "<strong>ref_thread_content:</strong> ".$row['ref_thread_content'].'<br />';
How do I get it to spit out every record in this table?
Thanks.
You'd need to use a while loop. the fetch functions only get one row at a time.
while($row = mysql_fetch_array($result)) {
echo ...
}
You're only grabbing the first record, loop through each one:
while ( $row = mysql_fetch_array( $result )) {
echo "<strong>ref_thread_id:</strong> ".$row['ref_thread_id'].'<br />';
echo "<strong>ref_thread_prefix:</strong> ".$row['ref_thread_prefix'].'<br />';
echo "<strong>ref_thread_topic:</strong> ".$row['ref_thread_topic'].'<br />';
echo "<strong>ref_thread_content:</strong> ".$row['ref_thread_content'].'<br />';
}
Use a loop:
$result = mysql_query("SELECT id, name FROM mytable");
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
printf("ID: %s Name: %s", $row[0], $row[1]);
}
(Example part taken from http://php.net/manual/en/function.mysql-fetch-array.php)
Related
<html><head></head><body>
<?php
$db_hostname = "mysql";
$db_database = "u1da";
$db_username = "u1da";
$db_password = "1234";
$con = mysql_connect($db_hostname ,$db_username ,$db_password);
if (!$con) die ("Unable to connect to MySQL: ".mysql_error ());
mysql_select_db ( $db_database ) ||
die (" Unable to select database : ". mysql_error());
$query = "select * from students";
$result = mysql_query($query);
$result || die ("Database access failed: ".mysql_error());
$rows = mysql_num_rows($result);
echo "<table border=1>";
echo "<tr><td><p><b>Name</b></p></td></tr>";
for ($j = 0; $j < $rows ; $j ++) {
echo "<tr><td>", mysql_result($result,$j,'name') ,"</td></tr>";
}
echo "</table><br />";
$query2 = "select * from groups";
$result2 = mysql_query($query2);
$result2 || die ("Database access failed: ".mysql_error());
$rows2 = mysql_num_rows($result2);
echo "<table border=1>";
echo "<tr><td><p><b>Tutorial Group</b></p></td><td><p><b>Capacity</b></p></td></tr>";
for ($i = 0; $i < $rows2 ; $i ++) {
echo "<tr><td>", mysql_result($result2,$i,'Tutorial_Group') ,"</td><td>", mysql_result($result2,$i,'Capacity') ,"</td></tr>";
}
echo "</table>";
echo "<form method='post' action='' enctype="multipart/form-data">";
$query3 = "select * from students";
$result3 = mysql_query($query3);
echo "<br /><br /><select name='name'>";
while($name = mysql_fetch_array($result3)) {
echo "<option value='$name[Name]' > $name[Name] </option>"."<BR>";
}
echo "</select><br />";
$query4 = "select Tutorial_Group from groups";
$result4 = mysql_query($query4);
echo "<select name = 'group'>";
while($grp = mysql_fetch_array($result4)){
echo "<option value='$grp[Tutorial_Group]'>$grp[Tutorial_Group]</option>";
}
echo "</select><br />";
echo "Student ID: ";
echo '<input type="text" name="SID"><br />';
echo "Email: ";
echo '<input type="text" name="email"><br />';
echo '<input type="submit" name="submit" value="Submit">';
echo "</form>";
?>
Here is my current script code.
I have to make a query which will get the value from the 2 drop-down menus and 2 text fields and insert them into a table.
Table is called assg and have columns: Name, Student_ID, email, s_group. I have tried some different ways, but it didn't worked out. Please help.
Your first problem is that you are using mysql_fetch_array to get an array of elements into $grp variable but next you try to use it with a associative array to get values like $grp[Tutorial_Group]
Using mysql_fetch_array you can get $grp[0] ... $grp[1] but not $grp[Tutorial_Group]
You need to use mysql_fetch_assoc to get associative arrays like $grp[Tutorial_Group]
Your second problem is using complex php vars incorrectly example
Incorrect:
echo "<option value='$grp[Tutorial_Group]'>$grp[Tutorial_Group]</option>";
Correct:
echo '<option value="'.$grp['Tutorial_Group'].'">'.$grp['Tutorial_Group'].'</option>';
Or
echo "<option value='{$grp['Tutorial_Group']}'>{$grp['Tutorial_Group']}</option>";
Also the code only has the part to view the tables and the form. The insert part of the data is missing in the example. Probably this part has also some errors.
No indentation, html in echos, i don't even understand what you want.
By dropdown you mean select ?
What about $_POST['nameOfTheField'] ?
And why multipart ? There's no files to send here.
I have a large database of venues - and I would like to display this data in one page that would only change in some sort of an attribute to the id: (Ex: venues.php?id=1, which would get all the data from row #1.)
Edit: Okay, I updated the code and this is what it looks like now:
<?php
mysql_connect("localhost", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".(int)$id) ;
or die(mysql_error());
Print "<table border cellpadding=3>";
while($info = mysql_fetch_array( $data ))
{
Print "<tr>";
Print "<th>Name:</th> <td>".$info['VENUE_NAME'] . "</td> ";
Print "<th>Address:</th> <td>".$info['ADDRESS'] . " </td></tr>";
}
Print "</table>";
?>
And upon going to venues.php?id=1 I get this error:
Parse error: syntax error, unexpected T_LOGICAL_OR in
/home/nightl7/public_html/demos/venues/venues.php on line 8
Do you mean something like:
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".(int)$id) ;
In order to "pass" the id into your url "venues.php?id=1"
You need to use a hybrid html/php form with method=get.
You can see an example html form here: w3schools html forms
This is what I would do:
print '<form name="input" action="venues.php" method="get">';
print 'Venue: <select name = "id">';
$con = mysql_connect("","","");
mysql_select_db($dataBase);
if (!$con){die('Could not connect: ' . mysql_error());}
else {
$opt = array();
$optVal = array();
$i = 0;
$sql = "Select * from venues";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result))
{
$opt[$i] = $row['VenueName'];
$optVal[$i] = $row['VenueID'];
print "<option value='$optVal[$i]'>$opt[$i]</option>";
$i++;
}
}
mysql_close($con);
print '</select><br />';
print '<input type="submit" value="Submit" />';
print '</form>'
This will give you a form that will give you a drop down list of all your venues and once a venue is selected will direct you to the venues.php page with the respective id.
at the top of your venues,php page just use
$id = $_GET['id'];
This assigns the id number to the variable $id and then you can use this "select"
$data = mysql_query("SELECT * FROM venues WHERE id = ".$id) ;
To collect your venue name from your database using the id supplied in the form.
Good Luck :)
<?php
mysql_connect("localhost", "", "") or die(mysql_error());
mysql_select_db("") or die(mysql_error());
$id = (int) $_GET['id'];
$data = mysql_query("SELECT * FROM venues WHERE id = ".$id) or die(mysql_error());
Print "<table border cellpadding=3>";
while($info = mysql_fetch_array( $data ))
{
Print "<tr>";
Print "<th>Name:</th> <td>".$info['VENUE_NAME'] . "</td> ";
Print "<th>Address:</th> <td>".$info['ADDRESS'] . " </td></tr>";
}
Print "</table>";
?>
My question is how to display all data from my users table in my database?
I have this.
$loop = mysql_query(“SHOW users FROM $dbname”) or die (‘cannot select tables’);
You want to SELECT the users, not SHOW them.
Basic SQL loop example:
$sql = mysql_query("SELECT * FROM `users`");
while ($row = mysql_fetch_object($sql)) {
echo $row->id . ' ' . $row->nickname . '<br />';
}
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
$result = mysql_query(“SELECT * FROM Users”, $link) or die (‘cannot select tables’);
while ($row = mysql_fetch_assoc($result)) {
echo $row['firstname'];
echo $row['lastname'];
echo $row['address'];
echo $row['age'];
}
mysql_free_result($result);
$loop = mysql_query('SELECT * FROM `users`') or die();
You don't want your database name to be in the query, and you want to be using SELECT
If you haven't connected to the database earlier then you need to add this before your query:
mysql_connect($mysql_host, $mysql_user, $user_password);
i'm trying to run this php code which should display a quote from mysql, but can't figure out where is it going wrong. the result variable is null or empty. can someone help me out. thanks!
<?php
include 'config.php';
// 'text' is the name of your table that contains
// the information you want to pull from
$rowcount = mysql_query("select count(*) as rows from quotes");
// Gets the total number of items pulled from database.
while ($row = mysql_fetch_assoc($rowcount))
{
$max = $row["rows"];
//print_r ($max);
}
// Selects an item's index at random
$rand = rand(1,$max)-1;
print_r ($rand);
$result = mysql_query("select * from quotes limit $rand, 1") or die ('Error: '.mysql_error());
if (!$result or mysql_num_rows($result))
{
echo "Empty";
}
else{
while ($row = mysql_fetch_array($result)) {
$randomOutput = $row['cQuotes'];
echo '<p>' . $randomOutput . '</p>';
}
}
$result = mysql_query("SELECT * FROM quotes ORDER BY rand() LIMIT 1") or die ('Error: '.mysql_error());
if (!$result || mysql_num_rows($result) == 0)
echo "Empty";
else {
while ($row = mysql_fetch_array($result)) {
$randomOutput = $row['cQuotes'];
echo '<p>' . $randomOutput . '</p>';
}
}
// your script probably can't go on without this file?
require 'config.php';
// I prefer to always pass the connection resource to mysql_query/mysql_real_escape_string
// assume $mysql = mysql_connect....
$result = mysql_query("SELECT Count(*) AS rows FROM quotes", $mysql)
or die(mysql_error());
// there's only one row with only one column, so mysql_result() is fine
$rowcount = mysql_result($result, 0, 0);
$rand = rand(0,$rowcount-1);
$result = mysql_query("SELECT cQuotes FROM quotes LIMIT $rand, 1", $mysql)
or die ('Error: '.mysql_error());
// there's either one or zero records. Again, no need for a while loop
$row = mysql_fetch_array($result, MYSQL_ASSOC);
if ( !$row ) {
echo "Empty";
}
else{
// do you have to treat $row['cQuotes'] with htmlspecialchars()?
echo '<p>', $row['cQuotes'], '</p>';
}
if ($result && mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
$randomOutput = $row['cQuotes'];
echo '<p>' . $randomOutput . '</p>';
}
} else {
echo "Empty";
}
$sqlQuery = "SELECT * FROM allowedUsers WHERE UserID = '" . $kUserID . "'";
$result=mysql_query($sqlQuery, $db);
if(!result)
{
echo "Error running query <br>" . mysql_error();
exit;
}
while($row = mysql_fetch_array($result))
{
echo $row[2];
}
I run the SQLQuery in phpMyAdmin and I am getting a valid result (1 row)
the table (allowedUsers) has 6 fields
I can't get anything out of the DB.
Any help is appreciated.
if(!result) should be if(!$result)
According to PHP.net's documentation, you don't need to pass $db to mysql_query(). Take a look at the example code:
<?php
mysql_connect("localhost", "mysql_user", "mysql_password") or
die("Could not connect: " . mysql_error());
mysql_select_db("mydb");
$result = mysql_query("SELECT id, name FROM mytable");
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
printf("ID: %s Name: %s", $row[0], $row[1]);
}
mysql_free_result($result);
?>
It may be helpful to see your connection code, ensure you've selected a database, etc.