Concatenate an echo statement having mysql_result - php

Question

I want to get the first name, last name and county to display but I am getting errors when I try to concatenate the echo statement here. How can I correct this?
$query = mysql_query("
SELECT `first_name`, `last_name`,`county`
FROM `contact`
WHERE `first_name` ='". mysql_real_escape_string(trim($_POST['name'])) ."'
");
echo (mysql_num_rows($query) !== 0)? mysql_result($query,0,'first_name'):'Name not found';
?>
This is the message I receive:
Warning: mysql_result(): last_name.county not found in MySQL result index 5 in C:\xampp\htdocs\ajax_Contact\name.php on line 14

Answer 1

This code should answer the OP's question in relation to his current code:
$query = mysql_query("
SELECT `first_name`, `last_name`,`county`
FROM `contact`
WHERE `first_name` ='". mysql_real_escape_string(trim($_POST['name'])) ."'
");
(mysql_num_rows($query) !== 0) ? print mysql_result($query,0,'first_name') : print 'Name not found';
?>
Also according to your error, make sure that this line:
`last_name`,`county`
is not currently
`last_name`.`county`
Notice the comma/period difference.
OR, if you are trying to display first name, last name, and county use this:
$query = mysql_query("
SELECT `first_name`, `last_name`,`county`
FROM `contact`
WHERE `first_name` ='". mysql_real_escape_string(trim($_POST['name'])) ."'
");
(mysql_num_rows($query) !== 0) ? print mysql_result($query,0,'first_name')." ".mysql_result($query,0,'last_name')." ".mysql_result($query,0,'county') : print 'Name not found';
?>
And ultimately for better performance and supported code use
$mysqli = new mysqli("your_host", "your_user", "your_password", "your_database");
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$query = "
SELECT `first_name`, `last_name`,`county`
FROM `contact`
WHERE `first_name` ='". mysql_real_escape_string(trim($_POST['name'])) ."'
";
$result = $mysqli->query($query);
while($row = $result->fetch_array()){
$rows[] = $row;
}
if($result->num_rows === 0){
print 'Name not found';
}
foreach($rows as $row){
print $row['first_name']." ".$row['last_name']." ".$row['county'];
}
/* free result set */
$result->close();
/* close connection */
$mysqli->close();

Answer 2

Your easyest way is to use a newer version of the MySQL functions - mysqli is a very good one to go with. But if you have to rely on the old function set, then pleace read the documentation; mysql_result returns the data, and you should prepare it yourself in the code.
Also, I have to agree with #MarcB - use mysql_fetch_assoc or something like that when oyu are only and really expecting one single row.
But if you have the time, merge to the mysqli or PDO methods - they are more modern (in other words they are OOP all the way) and safer.

Answer 3

First, don't use the mysql_*() library. It's deprecated and obsolete. But if you have to use it, then DON'T use mysql_result. It's inefficient. A better process is to fetch an entire row using the fetch functions
$result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_assoc($row);
if (mysql_num_rows($result) != 0) {
echo $row['first_name'], $row['last_name'], $row['county'];
}