sql query

[markl1988]

hi

i am trying to use an sql query which when you enter enter first or last name which excists in the table table, it should search the db and show the results id, first name, last name, and contratced_hoursmon_am fields from the record in database matching name entered on form and name field with a database record. however i am having huge problems executing this query, here is the code i am trying to use.

<?php // connects to database
require_once('Connections/mark.php');
error_reporting(E_ALL);
ini_set('display_errors', 'On');

if(isset($_POST['submit'])){
if(isset($_GET['go'])){
if(preg_match("/^[ a-zA-Z]+/", $_POST['name'])){
$name=$_POST['name'];
// get variables
if(isset($_POST['ID'])){
$ID = $_POST['ID'];}
// Open an sql query

$sql = "SELECT* ID, FirstName, LastName, contracted_hoursmon_am, FROM staffmember tbl".;
$result=mysql_query($query);
$numrows=mysql_num_rows($result);
// Loop through the results of the query
while($row=mysql_fetch_array($result)){
// Get the variable results
$ID = $conn->$row['ID'];
$FirstName = $conn->row['FirstName'];
$LastName = $conn->row['LastName'];
$contraced_hoursmon_am = $conn->row['contracted_hoursmon_am'];

while($row=mysql_fetch_assoc($result))

echo $FirstName=$row["FirstName"];
echo $LastName=$row["LastName"];
echo $ID=$row["ID"];
echo $contracted_hoursmon_am=$row["contracted_hoursmon_am"];
// print the variables fields retrieved from the record in database from the record with name enterd on form matching a name filed from a record in database
print_r($var);"$v'ID' <br>$v'FirstName' <br> $v'LastName' <br>$v'contracted_hoursmon_am' <br> ";
?>
<option value="<?=$ID?>" selected><?=($FirstName." ".$LastName)?></option>
<?php

}
// Closes the query

// Closes connection to database
?>

[garrettroyce]

$sql = "SELECT* ID, FirstName, LastName, contracted_hoursmon_am, FROM staffmember tbl"; // there's a . here that doesn't belong
$result=mysql_query($query);
//$numrows=mysql_num_rows($result); you should check if $result is valid before doing any operation on it
if (!$result) {
     exit(mysql_error());
}
if (mysql_num_rows($result) == 0) {
     exit('No rows found');
}

[markl1988]

now its saying i have a parse error on line 275, not even next to the code next to the </body> for some reason.

[garrettroyce]

You might be missing a ; } ) etc

[markl1988]

thanks

i have found the parse error and solved the other problem howver when i upload page to testing server it has no errors but when i enter a name and submit the form, it displays no reuslts at all for that member name entered on the form for some reason.

[xav0989]

echo $FirstName=$row["FirstName"];
echo $LastName=$row["LastName"];
echo $ID=$row["ID"];
echo $contracted_hoursmon_am=$row["contracted_hoursmon_am"];

It is quite normal that you don't see anything, as you are doing an echo not on the values, but on the result of the assignment.

[markl1988]

thanks

so what do u suggest to use to be able to display the reuslts other than the echo fucntion

[garrettroyce]

echo isn't the problem, the problem is PHP evaluates this:

echo $var = $array['index'];

in two steps:

$var = $array['index'];

echo ???

The ??? is what $var=$array['index'] is equal to. It's an expression, not a value. This would be what you want to do:

$var=$array['index'];
$var2=$array['index2'];
echo $var;
echo $var2;
echo 'text';
echo $var, $var2, 'text'; // or you can do it this way

[markl1988]

i tried that solution but still not displaying any records, i dunno what else could be done to solve this.

[garrettroyce]

Try changing your query to this:

$sql = "SELECT ID, FirstName, LastName, contracted_hoursmon_am, FROM staffmember";