Else if statements are giving me errors in PHP

[abrooks]

Please Please Please can somebody help me.
I have created a database and have a drop down menu within an html file and i am now using a php file to execute mysql queries from the drop down menu. everytime i use the }ifelse{ or }if command I keep getting an error. At the moment I can only use one query.

$query= "select Price, Area, Type from Properties where Price <100000";
$ mysql_result = mysql_query ($query,$mysql_link);
if($ViewMethod == "under £100,000");

if I try to add to this by doing

$query= "select Price, Area, Type from Properties where Price >100000 and Price >125000";
$ mysql_result = mysql_query ($query,$mysql_link);
if($ViewMethod == "£100,000-125000");

it doesn't work or if i put in an else it doesn't work. Can anyone out there help me with this.

[Drakla]

Can you post a snippet of your real code, as I get the gist of what you're saying, but it's a bit confused.

Plus you're not actually putting bits of the query in the dropdown, are you? If so that leaves you open to injection attacks if you're displaying it publically.

[abrooks]

Everything between if($ViewMethod == "under £100000"); and print "<table border=\"1\" width=\"100%\">"; is what i want to put in my code but can't

The other is the html.

I just want a drop down menu.

[abrooks]

the other file is this one:

[laserlight]

Basically,

if (condition1) {
	//some code
} elseif (condition2) {
	//some more code
} else {
	//yet more code
}

[abrooks]

Thank you for replying, but I have tried this several times:

<?php

$ mysql_link = mysql_connect ("mysql.lsbu.ac.uk", "brookeab", "once_fed");
$ testresult = mysql_select_db ( "brookeab" ,$mysql_link);

if($ViewMethod == "under £100000"){
$query= "select Price, Area, Type from Properties where Price <100000";

} else if($ViewMethod == "£100000-125000");
$query= "select Price, Area, Type from Properties where Price >100000 and Price >125000";

} else if($ViewMethod == "£125000-150000");
$query= "select Price, Area, Type from Properties where Price >125000 and Price >150000";

} else if($ViewMethod == "£150000-175000");
$query= "select Price, Area, Type from Properties where Price >150000 and Price >175000";
$ mysql_result = mysql_query ($query,$mysql_link);

print "<table border=\"1\" width=\"100%\">";

        while($row= mysql_fetch_row ($mysql_result)){
        print "<td width=\"10%\">$row[0]</td>";
        print "<td width=\"30%\">$row[1]</td>";
        print "<td width=\"30%\">$row[2]</td>";
        print "</tr>";
    }

  print "</table>";

?>

I keep getting an error
(Parse error: parse error in /users/cam/pbscst3/brookeab/.public_html/cgi-bin/process.php on line 26)

[Drakla]

Sorry dude, just drowned about 50 pikmin and completely forgot about coding.

There are stacks of errors in your code - first you can't have a space between the $ sign and the variable name, your query code was put into the last else if block of code, so would only run if that condition were met - plus the conditions had two greater thans in a lot of cases.

This version should work

<?php
$mysql_link = mysql_connect("mysql.lsbu.ac.uk", "brookeab", "once_fed");
$testresult = mysql_select_db( "brookeab" ,$mysql_link);
	// all the queries start the same - so set the guts of the query and tag on the conditions
$query = "SELECT Price, Area, Type FROM Properties WHERE ";

// could use switch instead of all these ifs

if($ViewMethod == "under £100000")
{
	$query.= "Price <100000";
}
else if($ViewMethod == "£100000-125000")
{
	$query.= "Price >100000 and Price <125000";
}
else if($ViewMethod == "£125000-150000")
{
	$query.= "Price >125000 and Price <150000";
}
else if($ViewMethod == "£150000-175000")
{
	$query.= "Price >150000 and Price <175000";
}

$mysql_result = mysql_query ($query, $mysql_link);

print '<table border="1" width="100%">';

while( $row = mysql_fetch_row($mysql_result) )
{
	print "<tr><td width=\"10%\">$row[0]</td>";
	print "<td width=\"30%\">$row[1]</td>";
	print "<td width=\"30%\">$row[2]</td>";
	print "</tr>";
}

print "</table>";
?>

Consider using BETWEEN on the queries, and as a small point always make table names and fields lower case as in the long run it'll save time if you never have to think about capitalisation.

Gah - just looked at the page with the dropdowns - the code for the select is wrong. Selects are like this

<select name="ViewOptions">
    <option value="10000-30000">Between 10 and 30 thousand</option>
        <option value="30000-80000">Between 30 and 80 thousand</option>
</select>

etcetera... and it's the value that gets passed for the box.

Plus unless you've got register globals on, which I recommend against, you'll have to suck the variable out of the $GET or $POST superglobal arrays, e.g.

$ViewMethod = $_POST['ViewMethod'];

[abrooks]

Thank you Drakla for the help but unfortnately it didn't work becasue I keep getting an error at:

while( $row = mysql_fetch_row($mysql_result) )

[Drakla]

change the query line to

$mysql_result = mysql_query($query, $mysql_link) or die(mysql_error() . '<br />Query = ' . $query);

And paste in what the error message is as it's probably something simple

[abrooks]

Thank you Drakla,
your last code did work but i had to do a bit of tweeking.
THANK YOU, THANK YOU, THANK YOU