I've searched about possible solutions to this. And I've tried trying the variants of mysql_fetch_assoc and mysql_fetch_array.
But the output stil: Resource id #5
Here is the code:
<?php
$sql = "select bidang.idRel from relationship, bidang where relationship.idRel = bidang.idRel";
$result = mysql_query($sql);
echo $result ;
?>
If i change it to
echo $result['idRel'] ;
Nothing shown.
What should I do?
Please help me.
You are print a resource, not a result set.
Results are attached to it, but, its not an actual result set.
You need to fetch the result set by looping over the result set.
E.g.
$sql = "select bidang.idRel from relationship, bidang where relationship.idRel = bidang.idRel";
$result = mysql_query($sql);
if (mysql_num_rows($result)) {
while($row = mysql_fetch_assoc($result)) {
print_r($row);
}
}
Note: Don't use mysql_ functions. They are deprecated and will be removed in future PHP versions. Use mysqli_ instead.
it should be like this
<?php
$sql = "select bidang.idRel from relationship, bidang where relationship.idRel = bidang.idRel";
$result = mysql_query($sql);
while($raw = mysql_fetch_array($result))
{
echo $raw['ColumnName'];
}
?>
Related
I am trying to query a db for an entire column of data, but can't seem to get back more than the first row.
What I have so far is:
$medicationItem = array();
$medicationItemSql = "SELECT medication FROM medication";
$medicationItemObj = mysqli_query($connection, $medicationItemSql);
if($row = mysqli_fetch_array($medicationItemObj, MYSQLI_NUM)){
echo count($row);
}
It's not my intention to just get the number of rows, I just have that there to see how many it was returning and it kept spitting out 1.
When I run the sql at cmd line I get back the full result. 6 items from 6 individual rows. Is mysqli_fetch_array() not designed to do this?
Well, I had a hard time understanding your question but i guess you are looking for this.
$medicationItem = array();
$medicationItemSql = "SELECT medication FROM medication";
$medicationItemObj = mysqli_query($connection, $medicationItemSql);
if($row = mysqli_num_rows($medicationItemObj))
{
echo $row;
}
Or
$medicationItem = array();
$medicationItemSql = "SELECT medication FROM medication";
$medicationItemObj = mysqli_query($connection, $medicationItemSql);
$i = 0;
while ($row = mysqli_fetch_array($medicationItemObj))
{
$medicationItem[] = $row[0];
$i++;
}
echo "Number of Rows: " . $i;
If you just want the number of rows i would suggest using the first method.
http://php.net/manual/en/mysqli-result.num-rows.php
You can wrote your code like below
$medicationItem = array();
$medicationItemSql = "SELECT medication FROM medication";
$medicationItemObj = mysqli_query($connection, $medicationItemSql);
while ($row = mysqli_fetch_assoc($medicationItemObj))
{
echo $row['medication'];
}
I think this you want
You could give this a try:
$results = mysqli_fetch_all($medicationItemObj, MYSQLI_NUM);
First, I would use the object oriented version of this and always use prepared statements!
//prepare SELECT statement
$medicationItemSQL=$connection->prepare("SELECT medication FROM medication");
// execute statement
$medicationItemSQL->execute();
//bind results to a variable
$medicationItemSQL->bind_result($medication);
//fetch data
$medicationItemSQL->fetch();
//close statement
$medicationItemSQL->close();
You can use mysqli_fetch_assoc() as below.
while ($row = mysqli_fetch_assoc($medicationItemObj)) {
echo $row['medication'];
}
please take a look at this code :
$sql = "SELECT * FROM shop";
$result = mysql_query($sql);
echo $result;
echo "before lop";
while ($xxx = mysql_fetch_assoc($result)) {
echo "inside lop";
echo $xxx['column_name'];
}
echo "after lop";
When I run such code i receive :
Resource id #244
before lop
after lop
It did not enter while lop, and I really don't know why :(
I used before such code and there were no problems.
Can someone help me?
$sql = "SELECT * FROM shop";
$result = mysql_query($sql) or die(mysql_error());
echo mysql_num_rows($result);
Check how many records are present in your shop table. I think shop table is empty.That is why not entering in the while loop.
You can do like this
$count = mysql_num_rows($result);
if($count > 0) {
while ($xxx = mysql_fetch_assoc($result)) {
echo $xxx['column_name'];
}
}
I would guess that the call to mysql_fetch_assoc() has returned false, possibly due to no results being returned from the database, this would cause the while loop to not execute even once. I would check the output of var_dump(mysql_fetch_assoc($result)) to ensure that data has been returned.
I have the following query:
$result = mysql_query("SELECT option_value FROM wp_10_options WHERE option_name='homepage'");
$row = mysql_fetch_array($result);
print_r ($row);
and the output I am getting is:
Resource id #2
Ultimately, I want to be able to echo out a single field like so:
$row['option_value']
Without having to use a while loop, as since I am only trying to get one field I do not see the point.
I have tried using mysql_result with no luck either.
Where am I going wrong?
Try with mysql_fetch_assoc .It will returns an associative array of strings that corresponds to the fetched row, or FALSE if there are no more rows. Furthermore, you have to add LIMIT 1 if you really expect single row.
$result = mysql_query("SELECT option_value FROM wp_10_options WHERE option_name='homepage' LIMIT 1");
$row = mysql_fetch_assoc($result);
echo $row['option_value'];
$result = mysql_query("SELECT option_value FROM wp_10_options WHERE option_name='homepage'");
$row = mysql_fetch_assoc($result);
echo $row['option_value'];
Functions mysql_ are not supported any longer and have been removed in PHP 7. You must use mysqli_ instead. However it's not recommended method now. You should consider PDO with better security solutions.
$result = mysqli_query($con, "SELECT option_value FROM wp_10_options WHERE option_name='homepage' LIMIT 1");
$row = mysqli_fetch_assoc($result);
echo $row['option_value'];
use mysql_fetch_assoc to fetch the result at an associated array instead of mysql_fetch_array which returns a numeric indexed array.
Though mysql_fetch_array will output numbers, its used to handle a large chunk.
To echo the content of the row, use
echo $row['option_value'];
Try this one if you want to pick only one option value.
$result = mysql_query("SELECT option_value FROM wp_10_options WHERE option_name='homepage'");
$row = mysql_fetch_array($result);
echo $row['option_value'];
What you should get as output with this code is:
Array ()
... this is exactly how you get just one row, you don't need a while loop. Are you sure you're printing the right variable?
Ultimately, I want to be able to echo out a signle field like so:
$row['option_value']
So why don't you? It should work.
It is working for me..
$show = mysql_query("SELECT data FROM wp_10_options WHERE
option_name='homepage' limit 1"); $row = mysql_fetch_assoc($show);
echo $row['data'];
is this is a WordPress?
You shouldn't do it like you've done!
To get option from DB use get_option!
this shoude work
<?php
require_once('connection.php');
//fetch table rows from mysql db
$sql = "select id,fname,lname,sms,phone from data";
$result = mysqli_query($conn, $sql) or die("Error in Selecting " . mysqli_error($conn));
//create an array
$emparray = array();
for ($i = 0; $i < 1; $i++) {
$row =mysqli_fetch_assoc($result);
} $emparray[] = $row;
echo $emparray ;
mysqli_close($connection);
?>
make sure your ftp transfers are in binary mode.
This is a part of my code, and the echo is to test the value and it gives me Resource ID #5
$id = mysql_query("SELECT id FROM users WHERE firstname='$submittedfirstname' AND lastname='$submittedlastname' AND email='$submittedemail'") or die(mysql_error());
$counter = mysql_num_rows($id);
echo $id;
I am just getting into programming, and lately seeing lot of Resource ID outputs/errors while working with Databases.
Can someone correct the error in my code? And explain me why it isnt giving me the required output?
This is not an error. This is similar to when you try to print an array without specifying an index, and only the string "Array" is printed. You can access the actual data contained within that resources (which you can think of as a collection of data) using functions like mysql_fetch_array().
In fact, if there were an error here, the value of $id would not be a resource. I usually use the is_resource() function to verify that everything is alright before using variables which are supposed to contain a resource.
I guess what you intend to do is this:
$result = mysql_query("SELECT id FROM users WHERE firstname='$submittedfirstname' AND lastname='$submittedlastname' AND email='$submittedemail'") or die(mysql_error());
if(is_resource($result) and mysql_num_rows($result)>0){
$row = mysql_fetch_array($result);
echo $row["id"];
}
Did you mean to echo $counter? $id is a resource because mysql_query() returns a resource.
If you are trying to get the value of the id column from the query, you want to use e.g., mysql_fetch_array().
Here is an excerpt from http://php.net/mysql.examples-basic:
$query = 'SELECT * FROM my_table';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
// Printing results in HTML
echo "<table>\n";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "\t<tr>\n";
foreach ($line as $col_value) {
echo "\t\t<td>$col_value</td>\n";
}
echo "\t</tr>\n";
}
echo "</table>\n";
Adapted to the code you provided, it might look something like this:
$result =
mysql_query("SELECT id FROM users WHERE firstname='$submittedfirstname' AND lastname='$submittedlastname' AND email='$submittedemail' LIMIT 1")
or die(mysql_error());
if( $row = mysql_fetch_array($result, MYSQL_ASSOC) )
{
$id = $row['id'];
}
else
{
// No records matched query.
}
Note in my code that I also added LIMIT 1 to the query, as it seems like you are only interested in fetching a single row.
are you looking for
while ($row = mysql_fetch_array($id)) {
echo $row['id'];
}
?
$kode_gel = substr($_GET['gel'],0,3);
$no_gel = substr($_GET['gel'],3,5);
$cek = mysql_query("SELECT id_arisan
FROM arisan WHERE kode_gel = '".$kode_gel."'
AND no_gel = '".$no_gel."'");
$result = mysql_fetch_array($cek);
$id = $result['id_arisan'];
header("location: ../angsuran1_admin.php?id=".$id);
I have a query which is designed to retireve the "name" field for all records in my "tiles" table but when I use print_r on the result all I get is the first record in the database. Below is the code that I have used.
$query = mysql_query("SELECT name FROM tiles");
$tiles = mysql_fetch_array($query);
I really cant see what I have done wrong, I have also tried multiple searches within google but I cant find anything useful on the matter at hand.
<?php
// Make a MySQL Connection
$query = "SELECT * FROM example";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo $row['name']. " - ". $row['age'];
echo "<br />";
}
?>
'mysql_fetch_array'
Returns an array that corresponds to the fetched row and moves the internal data pointer ahead.
This means that it returns array (contains values of each field) of A ROW (a record).
If you want other row, you call it again.
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
// Do something with $row
}
Hope this helps. :D
Use "mysql_fetch_assoc" instead of "mysql_fetch_array".
$query = mysql_query('SELECT * FROM example');
while($row = mysql_fetch_assoc($query)) :
echo $row['whatever'] . "<br />";
endwhile;
I believe you need to do a loop to invoke fetch array until it has retrieved all the rows.
while ($row = mysql_fetch_array($query) ) {
print_r( $row );
}