I am trying to show some text depending upon what a status is set to in a db table.
See my code below:
$result=mysql_query("SELECT * FROM hr_recruitment_stages where vacancy_ref='$vacancyref' order by added_on DESC limit 0,1")or die('ERROR 315' );
$row = mysql_fetch_array($result);
$stage_name = $row ['stage_name'];
if($stage_name['stage_name'] == 'Shortlisting') { echo"Shortlisting"; } else { echo"Not Shortlisting"; } ?>
However this doesnt seem to be working properly as it is showing as Not Shortlisting even when stage_name equals Shortlisting.
Any ideas why?
Its variable type mistake. Check your assigned variable, you assigned the Array Element not the entire array. so try like below.
<?php
$result = mysql_query("SELECT * FROM hr_recruitment_stages where vacancy_ref='$vacancyref' order by added_on DESC limit 0,1") or die('ERROR 315' );
$row = mysql_fetch_array($result);
$stage_name = $row['stage_name'];
if($stage_name == 'Shortlisting') {
echo"Shortlisting";
} else {
echo"Not Shortlisting";
}
?>
Refer this Article for PHP Array understanding.
http://php.net/manual/en/language.types.array.php
Related
Maybe I am overthinking this, but I have narrowed my query to find a row down to 1 result that I need, and it will not display. Wondering if someone could tell me what I am doing wrong.
$result = mysqli_query($link, "SELECT pageid FROM article ORDER BY id DESC LIMIT 1");
$row = mysqli_use_result($result);
echo $row;
I have it selecting the last row and supplying me with the stored data from the pageid of the last row.
I had to adapt my code. I believe it was because I use mysql. However, this code will work if you use mysqli
$pageid = "SELECT pageid FROM articles ORDER BY id DESC LIMIT 1";
$resultpageid = $link->query($pageid);
if ($resultpageid->num_rows > 0) {
while ($row = $resultpageid->fetch_assoc()) {
$pagenumber = $row["pageid"];
}
} else {
echo "0 results";
}
mysqli doesn't have any function to get a single column from a single row. You need to use one of the fetch methods e.g. fetch_array(). You don't need any loop if you use LIMIT 1.
Just fetch a single row and get the column from the returned array:
$pageid = "SELECT pageid FROM articles ORDER BY id DESC LIMIT 1";
$resultpageid = $link->query($pageid);
$row = $resultpageid->fetch_assoc();
// or
$row = $resultpageid->fetch_array();
if ($row) {
echo $row["pageid"];
} else {
echo "No record found!";
}
I have two php files. They each have a different query in them. Both of them work. Then I have one file where I include both files inside of it. The queries work on their individual pages, but in the page where they are both included only the first query works. They look something like this (The first one uses * because I pull out every row)
include 'connect.php';
$query = "SELECT * FROM table ORDER by jid DESC";
$ex = $mysqli->query($query) or die(mysqli_error());
$row_cnt = mysqli_num_rows($ex);
if ($row_cnt > 0) {
/* fetch associative array */
while ($row = $ex->fetch_assoc()) {
echo $row["one"] . $row["two"] . $row["three"] . $row["four"];
}
$result->free();
}
$mysqli->close($mysqli);
The second one is like this
include 'connect.php';
$q = "SELECT one, three FROM table ORDER by jid DESC";
$d = $mysqli->query($q) or die(mysqli_error());
$row_cnt = mysqli_num_rows($d);
if ($row_cnt > 0) {
/* fetch associative array */
while ($row = $d->fetch_assoc()) {
echo $row["one"] . $row["three"];
}
$result->free();
}
$mysqli->close($mysqli);
Then the file that includes them is just two includes like include ' ';
How do I get both queries to work on the one page?
add value in array, like this:
include 'connect.php';
$q = "SELECT one, three FROM jokes ORDER by jid DESC";
$d = $mysqli->query($q) or die(mysqli_error());
$row_cnt = mysqli_num_rows($d);
$values = array();
if ($row_cnt > 0) {
/* fetch associative array */
while ($row = $d->fetch_assoc()) {
$values[0][] = $row["one"];
$values[1][] = $row["two"];
$values[2][] = $row["three"];
$values[3][] = $row["four"];
}
$result->free();
}
$mysqli->close($mysqli);
... after, use foreach for print the value
I have not tested
How do I get both queries to work on the one page?
Just run them one after another like everyone does.
If something goes wrong - debug your code, like everyone does. Read all the error messages and act accordingly.
<?php
mysql_select_db($database_XXX, $XXX);
$result= mysql_query("SELECT COUNT(*) FROM news");
$total = mysql_result($result, 0, 0);
// create a random number
mt_srand((double)microtime()*1000000);
$number = mt_rand()%$total;
// get a random entry
$result= mysql_query("SELECT * FROM news LIMIT $number, 5");
$row = mysql_fetch_array($result);
?>
This is the PHP code I'm using and it pulls up random data from the table I want but I can't seem to figure out how to have it not show the current post it's on. Will I need to throw in an if statement in? If I do where would it go and how to impletment it. The only thing I can thing of is using if statements to check if the post_id on page matches the post_id posted. But I'm new to this and the only thing I can think of is.
if(!$row['post_id'] == $_GET['id']) {
}
I don't know what to make it do after. Also if anyone knows how or can help point me in the right direction that would be great. Thanks.
Here is the update of the total thing here. It still shows post it's already on. hope this helps. This is the php code for the page.
<?php
mysql_select_db($database_xxx, $xxx);
$result= mysql_query("SELECT COUNT(*) FROM news");
$total = mysql_result($result, 0, 0);
// create a random number
mt_srand((double)microtime()*1000000);
$number = mt_rand()%$total;
// get a random entry
$result= mysql_query(sprintf("SELECT * FROM news WHERE post_id <> %d LIMIT %d, 3", $post->post_id, $number));
$row = mysql_fetch_array($result);
?>
<?php
if (! isset($_GET['id']) || (int) $_GET['id'] === 0 ) {
echo "Incorrect input, aborting";
exit;
}
mysql_select_db($database_xxx, $xxx);
$sql = "SELECT * FROM news WHERE post_id = " . $_GET['id'];
// a line of debug to make sure things are as expected
$query = MYSQL_QUERY($sql);
// query your table for a match with post_id
if (mysql_num_rows($query) == "1")
// if a record is found, show the info
{
$fetch = mysql_fetch_array($query); // set $fetch to have the values from the table
} else {
echo "No match in database found."; // if no match is found, display this error
}
?>
Assuming that code is on the page where you view the main product:
$result= mysql_query(sprintf("SELECT * FROM news WHERE id <> %d LIMIT %d, 5", $post->id, $number));
Also, you should not be using mysql_* functions anymore as they are deprecated; checkout PDO
How could i get $result too equal nothing if the column doesn't exist in PHP?
I was thinking something like this:
$result=mysql_query("SELECT * FROM users WHERE username=$key DESC LIMIT 1")or die ($result = '');
But i was told thats wrong.
It's wrong because you're killing the script with die when a DB error occurs, rather than doing stuff when you find no row.
What you presumably need is more like:
$result = mysql_query($query);
if ($result) {
if ($row = mysql_fetch_assoc($result)) {
// do stuff with row
} else {
// do stuff without row
}
} else { // not needed but left here for illustration purposes
// this is the part that would occur, had you called mysql_query(...) or die;
die(mysql_error());
}
$result=mysql_query("SELECT * FROM users WHERE username= '$key' LIMIT 1")or die (mysql_error());
then check the result of mysql_num_rows()
If you mean that the result returns 0 rows, you can check with mysql_num_rows, like this:
$result=mysql_query("SELECT * FROM users WHERE username=$key DESC LIMIT 1");
if (mysql_num_rows($result) == 0)
$result = '';
Your code will set $result to '' if there's an error, in which case mysql_query returns false. It will also halt the code, since you're calling die(). An empty result set is not an error, however. In that case mysql_query returns a valid resource identifier with no rows. If I understand your question, this is what you want to do:
$result=mysql_query("SELECT * FROM users WHERE username=$key DESC LIMIT 1");
if (mysql_num_rows($result) == 0){
$result = '';
}
<?php
// Here I assume you're using PHP PDO
$pdo = new PDO("mysql:server=localhost;dbname=mydatabase", "root", "");
$result = $pdo->query("SELECT * FROM users WHERE username=$key DESC LIMIT 1");
$errorcode = $pdo->errorCode();
$errorinfo = $pdo->errorInfo();
// Columns doesn't exist
if($errorcode == "43072") $result = "";
// Other error...
else if($errorcode != "00000") die("MySQL Error: " . $errorinfo[2]);
?>
I have the following inside a foreach loop (displaying my various videos), I'm trying to display some alternate text for the top three voted videos. What on earth am I doing wrong (a lot clearly)...
$sql = "SELECT video_id FROM videos WHERE displayable='y' ORDER BY votes desc LIMIT 0,3";
$result = mysql_query($sql);
$row = #mysql_fetch_array($result);
if(in_array($video->getProperty('video_id')) == $row['video_id']) {
do this...
} else {
do this..
}
Firstly replace your code with some error preventing techniques like so!
$sql = "SELECT video_id FROM videos WHERE displayable='y' ORDER BY votes desc LIMIT 0,3";
if(false != ($result = mysql_query($sql))
{
$row = mysql_fetch_assoc($result); //Dont need the # restraint as the result is not false above.
//Also to get associate keys you need to use mysql_fetch_assoc
if($video->getProperty('video_id') == $row['video_id'])) //Remove the in array as your directly comparing the to entities with ==
{
//Match
}else
{
//Video does not match
}
}
Your main problem was the mysql_fetch_array(), Please research the differences with mysql_fetch_array() and mysql_fetch_assoc();
--
Edit: The way i would go
//Change the query and the loop way.
$sql = "SELECT video_id FROM videos WHERE displayable='y' AND video_id != '".(int)$video->getProperty('video_id')."' ORDER BY votes desc LIMIT 0,3";
if(false != ($result = mysql_query($sql))
//Use the # restraint if you have E_NOTICE on within E_Reporting
{
while($row = mysql_fetch_assoc($result))
{
//Print the $row here how you wish for it to be displayed
}
}else
{
//We have an error?
echo '<strong>Unable to list top rated videos, please check back later.</strong>'
}
}
mysql_fetch_array only returns a single row, you need to loop through your results to build an array containing the top three ids.
$sql = "SELECT video_id FROM videos WHERE displayable='y' ORDER BY votes desc LIMIT 0,3";
$result = mysql_query($sql);
while($row = #mysql_fetch_array($result)) {
$topthree[] = $row["video_id"];
}
Then you can use in_array but with the correct syntax:
if(in_array($video->getProperty('video_id'), $topthree)) {
do this...
} else {
do this..
}