Echo items with value of three - php

Question

I need to echo all items in my column that have the value of three. If a field has the value of three I then need to echo the 'name' and 'description' on that row.
This is what I have so far
$result = mysql_query($query1) or die(mysql_error());
while ($row = mysql_fetch_array($result))
{
echo $row['status'];
}
I need to write 'if $row[`status´ == 3 echo 'description' 'name' else echo 'no current staus'
I hope I made some sense because I am seriously confused

Answer 1

You can actually select the result which only has 3 value in its status
for that use this
$query = "SELECT * FROM yourtable WHERE status='3'";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array(result)) {
echo "Name: ".$row['name']."<br>Description: ".$row['description'];
}
and if you want to parse you result to display only those of value this then use this
$query = "SELECT * FROM yourtable";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array(result)) {
if($row['status']==3) {
echo "Name: ".$row['name']."<br>Description: ".$row['description'];
}
}

Answer 2

if $row['status'] == 3 {
echo ($row['description']." ".$row['name']);
}
else
{
echo ('no current staus');
}
The "." in the first echo means string concatenation. I'm separating description and name with a single space character, feel free to change this.

Answer 3

I'm expecting something more since you have the "i need to write" part that is the answer?
$result = mysql_query($query1) or die(mysql_error());
while ($row = mysql_fetch_assoc($result)) // fetch_assoc better for your case
{
if( $row['status'] == 3 ) {
echo $row['description'];
}else {
echo 'no current staus';
}
}
For the sake of info for tasks with a numeric status
switch( $row['status'] ) {
case 0:
// do something
break;
case 1:
// do something
break;
case 2:
// do something
break;
case 3:
// do something
break;
default:
echo 'No status set';
}

Answer 4

if ($row['status'] == 3) {
echo $row['name'];
echo $row['description'];
}

Answer 5

$result = mysql_query($query1) or die(mysql_error());
while ($row = mysql_fetch_array($result))
{
if ($row['status'] == 3) {
echo $row['name'] . ' - ' . $row['description'];
}
else {
echo 'no curent status';
}
}

Answer 6

if($row['status']==3) {
echo $row['description'];
echo $row['name'];
} else {
echo 'no current status';
}
Maybe you should learn about basics of PHP and algorithms and make sure you understand this, before you move on.

Answer 7

This?
$result = mysql_query($query1) or die(mysql_error());
while ($row = mysql_fetch_array($result))
{
if (!in_array(3, $row['status']))
{
echo 'no current staus';
break;
}
echo $row['description'] . <br />;
echo $row['name'] . <br />;
}

Answer 8

or you could use SQL to select the values for you...
$query1 = SELECT * FROM table WHERE column = '3';
$result = mysql_query($query1) or die(mysql_error());
while ($row = mysql_fetch_array($result))
{
echo $row['status'];
}