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**fixed** This is making me nuts... error near " on line 1???

jmilane

By jmilane
in PHP Coding Help

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jmilane Newbie

jmilane

Posted
Posted
I have been at this for 3 hours... am I missing something obvious???

[code]<html>
<head>
<title>
ULEM Registration Database Search Form
</title>
<body>

<?PHP

$username="ghf";
$password="gfghgfh";
$database="ghgfh";
$databox="www.ghfhgfh.org";

mysql_connect($databox,$username,$password);

@mysql_select_db($database) or die("Unable to select database");

if (isset($_POST['regid'])) {
$regid = $_POST['regid'];
$query = "select employmentprog, businessprog, certprog, youthprog, parentprog, policyprog, firstname, middlename, lastname, birthdate, street, city, state, postalcode, homephone, otherphone, referral, contact, race, otherrace, gender, regdate from registration where id = $regid";
}

if (isset($_POST['lastname'])) {
$ln = $_POST['lastname'];
$query = "select employmentprog, businessprog, certprog, youthprog, parentprog, policyprog, firstname, middlename, lastname, birthdate, street, city, state, postalcode, homephone, otherphone, referral, contact, race, otherrace, gender, regdate from registration where lastname = $ln";
}

$result=mysql_query($query) or die(mysql_error()); 

$num=mysql_num_rows($result);

if ($num == 0) {echo "No matching records."; return;}

mysql_close();

echo "<b><center>The results of your search:</center></b><br><br>";

$i=0;
while ($i < $num) {

$firstname=mysql_result($result,$i,"firstname");
$middlename=mysql_result($result,$i,"middlename");
$lastname=mysql_result($result,$i,"lastname");
$homephone=mysql_result($result,$i,"homephone");
$otherphone=mysql_result($result,$i,"otherphone");
$street=mysql_result($result,$i,"street");
$city=mysql_result($result,$i,"city");
$state=mysql_result($result,$i,"state");
$postalcode=mysql_result($result,$i,"postalcode");
$referral=mysql_result($result,$i,"referral");
$race=mysql_result($result,$i,"race");
$otherrace=mysql_result($result,$i,"otherrace");
$contact=mysql_result($result,$i,"contact");
$gender=mysql_result($result,$i,"gender");
$regdate=mysql_result($result,$i,"regdate");

echo "<b>$firstname $lastname</b><br>Phone: $homephone<br>Other Phone: $otherphone<br>Street: $street<br>City: $city<br>Zip: $postalcode<br>Best time to contact $contact<br>Race: $race<br>Other race (if applicable): $otherrace<br>Gender: $gender<br>Registration date: $regdate<br><hr>";

$i++;
}

?>
</html>[/code]
wildteen88 Regular Member

wildteen88

Posted
Posted
Could you post the full error message(s) here please as it'll help us understand whats going on. I think the error you are getting isnt to do with PHP but to do with MySQL queries. As the "[i]error near .... on line 1[/i]" is common with errors in SQL queries as the error is comming from the database rather than PHP itself.
kenrbnsn Regular Member

kenrbnsn

Posted
Posted
[code]Change this line:
<?php $result=mysql_query($query) or die(mysql_error());  ?>
to
<?php $result=mysql_query($query) or die("Problem with the query: $query<br>" . mysql_error());  ?>[/code]
This will tell you if the query has a problem (which is usually the case when you get an error like that)

BTW, it looks like you're getting all the fields from your data base. If this is the case I would write your query as:
[code]<?php
$whr = '';
if (isset($_POST['regid'])) $whr = " where id = '" . mysql_real_escape_string($_POST['regid']) . "'";
if (isset($_POST['lastname'])) $whr = " where ln = '" . mysql_real_escape_string($_POST['lastname']) . "'";
$query = "select * from registration" . $whr;
?>[/code]

Ken

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