[SOLVED] Unexpected '='

[CrownVictoriaCop]

Hello again guys,

 

I'm using this code to retrieve information from a MySQL Database.

<?php
$username="safetyfi_secure";
$password="HIDDEN";
$database="safetyfi_students";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM tablename";
$result=mysql_query($query);

$num=mysql_numrows($result);

mysql_close();

echo "<b><center>Database Output</center></b><br><br>";

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

$field1-name = mysql_result($result,$i,"ID");
$field2-name = mysql_result($result,$i,"ParentFirstName");
$field3-name = mysql_result($result,$i,"ParentLastName");
$field4-name = mysql_result($result,$i,"StudentName");
$field5-name = mysql_result($result,$i,"Class");
$field6-name = mysql_result($result,$i,"Address");
$field7-name = mysql_result($result,$i,"City");
$field8-name = mysql_result($result,$i,"State");
$field9-name = mysql_result($result,$i,"Zip");
$field10-name = mysql_result($result,$i,"Phone");
$field11-name = mysql_result($result,$i,"BirthMonth");
$field12-name = mysql_result($result,$i,"BirthDay");
$field13-name = mysql_result($result,$i,"BirthYear");
$field14-name = mysql_result($result,$i,"School");
$field15-name = mysql_result($result,$i,"PaymentMethod");

echo "Student
ID: $field1-name<p>Enrolled in Class: $field5-name</p>
<p>$field4-name $field3-name</p>
<p>$field14-name</p>
<p>$field6-name</p>
<p>$field7-name, WI $field9-name</p>
<p>$field10-name</p>
<p>Parent Name: $field2-name $field3-name</p>
<p>Payment Method: $field15-name</p>
<hr>";

$i++;
}

?>

 

When using it, I'm getting the error

 

Parse error: syntax error, unexpected '=' in /home2/safetyfi/public_html/secure/directory.php on line 20

 

Can anyone tell me how I could fix this?

 

Thanks,

 

Anthony

[smerny]

try changing $field1-name to $field1_name, etc

[CrownVictoriaCop]

Fixes the error, but now no info from the database is being displayed.

try changing $field1-name to $field1_name, etc

[smerny]

you changed the variable names for the output as well, correct? what are you seeing exactly?

[CrownVictoriaCop]

The Words "Database Output". I don't think I did.

 

I changed some things in the script in hopes of making it working.

 

<?php
$username="safetyfi_secure";
$password="Hidden";
$database="safetyfi_students";

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM Students";
$result=mysql_query($query);

mysql_close();

echo "<b><center>Database Output</center></b><br><br>";

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

$field1_name = mysql_result($result,$i,"ID");
$field2_name = mysql_result($result,$i,"ParentFirstName");
$field3_name = mysql_result($result,$i,"ParentLastName");
$field4_name = mysql_result($result,$i,"StudentName");
$field5_name = mysql_result($result,$i,"Class");
$field6_name = mysql_result($result,$i,"Address");
$field7_name = mysql_result($result,$i,"City");
$field8_name = mysql_result($result,$i,"State");
$field9_name = mysql_result($result,$i,"Zip");
$field10_name = mysql_result($result,$i,"Phone");
$field11_name = mysql_result($result,$i,"BirthMonth");
$field12_name = mysql_result($result,$i,"BirthDay");
$field13_name = mysql_result($result,$i,"BirthYear");
$field14_name = mysql_result($result,$i,"School");
$field15_name = mysql_result($result,$i,"PaymentMethod");

echo "Student
ID: $field1_name<br>Enrolled in Class: $field5_name
<br>$field4_name $field3_name
<br>$field14_name
<br>$field6_name
<br>$field7_name, WI $field9_name
<br>$field10_name
<br>Parent Name: $field2_name $field3_name
<br>Payment Method: $field15_name
<hr>";

$i++;
}

?>

[Zane]

mysql_numrows should be mysql_num_rows

[CrownVictoriaCop]

mysql_numrows should be mysql_num_rows

I got rid of that part, simply because it displays the number of rows.

[Zane]

well that part was also running your loop

$num=mysql_numrows($result);

[...]

while ($i

 

now you don't have a $num to be greater than $i

 

Also..there is a

 tag for displaying code

[CrownVictoriaCop]

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home2/safetyfi/public_html/secure/directory.php on line 11

 

That's the error that comes up now.

well that part was also running your loop

$num=mysql_numrows($result);

[...]

while ($i < $num) {

 

now you don't have a $num to be greater than $i

 

Also..there is a

 tag for displaying code

[Zane]

When you get that error it means you need to do this

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

 

it will tell you why $result is not a valid mysql result resource

[CrownVictoriaCop]

When you get that error it means you need to do this

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

 

it will tell you why $result is not a valid mysql result resource

Found out the error and fixed it. Thank you guys again for helping a n00b!