[SOLVED] Help with if() statement

[Trium918]

I want to be able to select all user if All Users

are selected from the drop down menu.

 

Example: Click Here!!

 

$searchtype = ($_POST['searchtype']);
$all_users = ($_POST['all_users']);

$query="SELECT * FROM people where sponsor='$searchtype'";

echo "<table border=1 width=50% align=center>
<tr>
<td align=center>Name</td>
<td align=center>Age</td>
<td align=center>State</td>
<td align=center>Sponsor</td>
</tr>";

$result = mysql_query($query);
$num_results = mysql_num_rows($result); 
    
for ($i = 0; $i < $num_results; $i++) {
$row = mysql_fetch_array($result);
  
$name = htmlspecialchars( stripslashes($row["name"]));
$age = htmlspecialchars( stripslashes($row["age"]));
$state = htmlspecialchars( stripslashes($row["state"]));
$sponsor = htmlspecialchars( stripslashes($row["sponsor"]));

// Output
echo " <tr>
<td>$name</td>
<td>$age</td>
<td>$state</td>
<td>$sponsor</td>";

}// End of For loop
   
echo "</tr>
</table>";

if($searchtype==$all_users){
  $query2 = "SELECT * FROM people";
}

$query2 = "SELECT DISTINCT sponsor FROM people";
$result2 = mysql_query($query2);
$num_results2 = mysql_num_rows($result2);

echo "<form name=\"auto\" method=post>";
echo "<select name=\"searchtype\" onChange=\"auto.submit();\">
<option SELECTED>Sponsor</option>";
echo "<option name=\"all_users\">All Users</option>";
for ($i = 0; $i < $num_results2; $i++) {
    $row = mysql_fetch_array($result2);
    echo "<option value=\"$row[sponsor]\">$row[sponsor]</option>\n";
} 
echo "</select>";
//echo "<input type=\"submit\" value=\"Submit\" />";
echo "</form>";

[Trium918]

I'm sorry for those who have viewed this topic

for not having the link working. It is now up and

running.

 

Example: Click Here:

[papaface]

You don't need this:

$all_users = ($_POST['all_users']);

and change:

if($searchtype==$all_users){

to:

if($searchtype== "all_users"){

 

See what you get.

[Trium918]

You don't need this:

$all_users = ($_POST['all_users']);

and change:

if($searchtype==$all_users){

to:

if($searchtype== "all_users"){

 

See what you get.

 

Table didn't populate.

[Trium918]

Any Ideas

[sasa]

try

$searchtype = ($_POST['searchtype']);
$all_users = ($_POST['all_users']);
if($all_useers='all_users') $x = '1'; else $x="sponsor='$searchtype'";
$query="SELECT * FROM people where ".$x;
etc.

[Trium918]

try

$searchtype = ($_POST['searchtype']);
$all_users = ($_POST['all_users']);
if($all_useers='all_users') $x = '1'; else $x="sponsor='$searchtype'";
$query="SELECT * FROM people where ".$x;
etc.

 

I couldn't get it working. Here is what I tried.

 

if($all_user=$searchtype){ $query="SELECT * FROM people";}

else{echo "Could not run query";}

[Trium918]

Can someone please help me out?

 

[trq]

Can someone please help me out?

 

You have failed to ask an actual question.

[Trium918]

I was wondering about wether or not I was

on the right path with this. I am trying to populate

a table wiht every person in the database if $all_users

is selected in the drop down menu. Please read comments

in code where it says //COMMENTS?

 

Example; Click Here!!

 

$searchtype = ($_POST['searchtype']);	
$all_users = ($_POST['all_users']);

//COMMENTS
/* Ok, if $searchtype == something
then select all from database. $query="SELECT * FROM people";
is not returning true because it is skipped. Why*/

if($searchtype == "$all_users"){
     $query="SELECT * FROM people";
}else{
$query="SELECT * FROM people where sponsor='$searchtype'";}

echo "<table border=1 width=50% align=center>
<tr>
<td >Name</td>
<td >Age</td>
<td >State</td>
<td >Sponsor</td>
</tr>";

$result = mysql_query($query);
$num_results = mysql_num_rows($result); 
    
for ($i = 0; $i < $num_results; $i++) {
$row = mysql_fetch_array($result);
  
$name = htmlspecialchars( stripslashes($row["name"]));
$age = htmlspecialchars( stripslashes($row["age"]));
$state = htmlspecialchars( stripslashes($row["state"]));
$sponsor = htmlspecialchars( stripslashes($row["sponsor"]));

// Output
echo " <tr>
<td>$name</td>
<td>$age</td>
<td>$state</td>
<td>$sponsor</td>";

}// End of For loop
   
echo "</tr>
</table>";

$query2 = "SELECT DISTINCT sponsor FROM people";
$result2 = mysql_query($query2);
$num_results2 = mysql_num_rows($result2);

echo "<form name=\"auto\" method=post>";
echo "<select name=\"searchtype\" onChange=\"auto.submit();\">
<option SELECTED>Sponsor</option>";

 

Rest of code!!

//COMMENTS
/* I need a value from this line
so why isn't it returning a value?

echo "<option name=\"all_users\" >All Users</option>";

for ($i = 0; $i < $num_results2; $i++) {
$row = mysql_fetch_array($result2);
echo "<option value=\"$row[sponsor]\">$row[sponsor]</option>\n";
} 
echo "</select>";
//echo "<input type=\"submit\" value=\"Submit\" />";
echo "</form>";

 

[trq]

Looks good to me.

[Trium918]

I was wondering about wether or not I was

on the right path with this. I am trying to populate

a table wiht every person in the database if $all_users

is selected in the drop down menu. Please read comments

in code where it says //COMMENTS?

 

Example; Click Here!!

 

<?php
$searchtype = ($_POST['searchtype']);	
$all_users = ($_POST['all_users']);

//COMMENTS
/* Ok, if $searchtype == something
then select all from database. $query="SELECT * FROM people";
is not returning true because it is skipped. Why*/

if($searchtype == "$all_users"){
     $query="SELECT * FROM people";
}else{
$query="SELECT * FROM people where sponsor='$searchtype'";}

echo "<table border=1 width=50% align=center>
<tr>
<td >Name</td>
<td >Age</td>
<td >State</td>
<td >Sponsor</td>
</tr>";

$result = mysql_query($query);
$num_results = mysql_num_rows($result); 
    
for ($i = 0; $i < $num_results; $i++) {
$row = mysql_fetch_array($result);
  
$name = htmlspecialchars( stripslashes($row["name"]));
$age = htmlspecialchars( stripslashes($row["age"]));
$state = htmlspecialchars( stripslashes($row["state"]));
$sponsor = htmlspecialchars( stripslashes($row["sponsor"]));

// Output
echo " <tr>
<td>$name</td>
<td>$age</td>
<td>$state</td>
<td>$sponsor</td>";

}// End of For loop
   
echo "</tr>
</table>";

$query2 = "SELECT DISTINCT sponsor FROM people";
$result2 = mysql_query($query2);
$num_results2 = mysql_num_rows($result2);

echo "<form name=\"auto\" method=post>";
echo "<select name=\"searchtype\" onChange=\"auto.submit();\">
<option SELECTED>Sponsor</option>";
?>

 

Rest of code!!

//COMMENTS
/* I need a value from this line
so why isn't it returning a value?

echo "<option name=\"all_users\" >All Users</option>";

for ($i = 0; $i < $num_results2; $i++) {
$row = mysql_fetch_array($result2);
echo "<option value=\"$row[sponsor]\">$row[sponsor]</option>\n";
} 
echo "</select>";
//echo "<input type=\"submit\" value=\"Submit\" />";
echo "</form>";

 

[Trium918]

Does anyone have any Ideas?

[freakus_maximus]

if($searchtype == "$all_users"){

 

On this line, your IF is never going to be true. Remove the quotes you have around $all_users.

[Barand]

/* I need a value from this line

so why isn't it returning a value?

 

echo "<option name=\"all_users\" >All Users</option>";

 

You haven't specified a value, you specified a name

 

Change it to

 

echo "<option value='all_users' >All Users</option>";

[Trium918]

Would I use isset($searchtype) because I don't

want it to automatically populate the table.

 

Example: Click Here!!

if($searchtype == "$all_users"){
	$query="SELECT * FROM people";
}else{
$query="SELECT * FROM people where sponsor='$searchtype'";}


             echo "<option value=\"$all_users\" >All Users</option>";

[Barand]

isset($searchtype) will always return "true" no matter what value it has, so you need to check its value as you have in your code ^

[Trium918]

isset($searchtype) will always return "true" no matter what value it has, so you need to check its value as you have in your code ^

 

Ok, even when I take out $all_users it's populating

the table still. That means that if($searchtype == $all_users)

is returning true still, why?

 

echo "<option value=\"\" >All Users</option>";

echo "<option value=\"$all_users\" >All Users</option>";

[kenrbnsn]

You want:

<?php
echo '<option value="all_users" >All Users</option>';
?>

Then the "if" statement

<?php
if ($searchtype == 'all_users')
?>

would work.

 

Ken

[Trium918]

You want:

<?php
echo '<option value="all_users" >All Users</option>';
?>

Then the "if" statement

<?php
if ($searchtype == 'all_users')
?>

would work.

 

Ken

 

Ken it worked but what is the differents between the two? Thanks!

echo '<option value="all_users" >All Users</option>';

echo "<option value="all_users" >All Users</option>";