[SOLVED] Drop Down Menu

[Trium918]

I need help selecting all the the information

from the database where the selected items

are chose from the drop down menu.

 

<?
$strServer="localhost"; // Server IP Address 'or' Name
$strDatabase="mylabser_cms"; // Database Name

$strDB=mysql_connect($strServer) or die("Connection to database failed");
$database=mysql_select_db($strDatabase) or die("Database selection Failed");

//$query="Select * FROM people";
$query="Select * FROM people WHERE".searchtype; //the problem is here but I dont know
                                                                      //the syntax

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);

if ($searchtype) {
for ($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>";
?>

 

<form method="post">
    Choose Search Type:<br>
    <select name="searchtype">
      <option value="name">Name  
      <option value="age">Age
      <option value="state">State
   <option value="sponsor">Sponsor
    </select>
    <br /><br />
    <input type=submit value="Search">
  </form>

[papaface]

try:

<?
$strServer="localhost"; // Server IP Address 'or' Name
$strDatabase="mylabser_cms"; // Database Name




$strDB=mysql_connect($strServer) or die("Connection to database failed");
$database=mysql_select_db($strDatabase) or die("Database selection Failed");


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

//$query="Select * FROM people";
$query="Select * FROM people WHERE type='{$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);




if ($searchtype) {
for ($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>";
?>

You chould change:

$query="Select * FROM people WHERE type='{$searchtype}'";

 

type, to the column that you want to use the where clause on.

[MadTechie]

try

$searchtype = mysql_real_escape_string($_POST['searchtype']);
$query="Select * FROM people WHERE type=".searchtype;

 

D'oh too slow

[Trium918]

It donnot work. I'm not getting an error but it will

not pull anything from the database. Maybe it has

something to do with my form.

 

<form method="post">
    Choose Search Type:<br>
    <select name="searchtype">
      <option value="name">Name  
      <option value="age">Age
      <option value="state">State
   <option value="sponsor">Sponsor
    </select>
    <br /><br />
    <input type=submit value="Search">
  </form>

 

This works but I trying to get the drop down menu

to work.

$query="Select * FROM people"

[Barand]

It looks to me as though $searchtype contains the column to be searched. But where is the value you are searching for?

[Trium918]

Shouldn't the value = "name" attribute contain the value.

[Barand]

Add another text box to your form

Search for <input type='text' name='searchvalue' size='20'>

 

Then

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

$query="Select * FROM people WHERE $searchtype = '$searchvalue' ";

?>

 

[Trium918]

I'm only trying to get the drop down menu

to work.

[Barand]

Getting it to work is one thing. Getting it to do something useful when it does work  is another.

 

I apologise for trying to help. It won't happen again.

[shocker-z]

*slaps barands hands*  Your just too helpful you are mate!!

 

 

*tut tut*

 

[Barand]

There are those you can help and those you can't. As far as I'm concerned, this guy's on his own from now on.

 

[Trium918]

You people are rude. I would never give up on

no one. I would always help until they understand.

I cannont help that I'm not as skillful as you at php.

Im trying!!

[Barand]

On the contrary, I offered help and you threw it back in my face. And I did apologize.

[Trium918]

I have this book where it uses the same method

you gave me. All I was asking is there away that

I can get it woking without adding a extra text field?

 

<html>
<head>
  <title>Book-O-Rama Search Results</title>
</head>
<body>
<h1>Book-O-Rama Search Results</h1>

 

<?
  if (!$searchtype || !$searchterm)
  {
     echo "You have not entered search details.  Please go back and try again.";
     exit;
  }
  
  $searchtype = addslashes($searchtype);
  $searchterm = addslashes($searchterm);

  @ $db = mysql_pconnect("localhost", "bookorama", "bookorama");

  if (!$db)
  {
     echo "Error: Could not connect to database.  Please try again later.";
     exit;
  }

  mysql_select_db("books");
  $query = "select * from books where ".$searchtype." like '%".$searchterm."%'";
  $result = mysql_query($query);

  $num_results = mysql_num_rows($result);

  echo "<p>Number of books found: ".$num_results."</p>";

  for ($i=0; $i <$num_results; $i++)
  {
     $row = mysql_fetch_array($result);
     echo "<p><strong>".($i+1).". Title: ";
     echo stripslashes($row["title"]);
     echo "</strong><br>Author: ";
     echo stripslashes($row["author"]);
     echo "<br>ISBN: ";
     echo stripslashes($row["isbn"]);
     echo "<br>Price: ";
     echo stripslashes($row["price"]);
     echo "</p>";
  }

?>

</body>
</html>

 

 

[Trium918]

The drop down menu is being populated, but only

when the first For loop is commented out. How can I

fix it where the menu is populated and able to pull data

from the database. I need the drop down menu to populate

the table inside the first commented For loop.

 

<?
$strServer="localhost"; // Server IP Address 'or' Name
$strDatabase="mylabser_cms"; // Database Name

$strDB=mysql_connect($strServer) or die("Connection to database failed");
$database=mysql_select_db($strDatabase) or die("Database selection Failed");
$query="Select sponsor FROM people";

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++) {
  
$name = mysql_result($result,$i,"name");
$age = mysql_result($result,$i,"age");
$state = mysql_result($result,$i,"state");
$sponsor = mysql_result($result,$i,"sponsor");

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

      }// End of For loop

	      echo "</tr>
     </table>";*/

// Calling Function For Drop Down Menu
   Populating_DDM();

function Populating_DDM(){
global $num_results,$result;
echo "<select value=''>Sponsor</option>";
// printing the list box select command

for ($i = 0; $i < $num_results; $i++) {
$row = mysql_fetch_array($result);
echo "<option value=$row[$i]>$row[sponsor]</option>";

/* Option values are added by looping through the array */
}
echo "</select>";// Closing of list box 
}// End of function
?>

 

[MadTechie]

i have no idea what your trying to do.. did you skip a chapter as

 

$name = mysql_result($result,$i,"name");
$age = mysql_result($result,$i,"age");
$state = mysql_result($result,$i,"state");
$sponsor = mysql_result($result,$i,"sponsor");

 

isn't going to work..

 

 

[Trium918]

i have no idea what your trying to do.. did you skip a chapter as

 

$name = mysql_result($result,$i,"name");
$age = mysql_result($result,$i,"age");
$state = mysql_result($result,$i,"state");
$sponsor = mysql_result($result,$i,"sponsor");

 

isn't going to work..

 

It is pulling data from the database and populating

the tables. I also have this:

 

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

 

Both are working.

[MadTechie]

that will work.. i still don't know what the problem is .. or what your goal is..

[Trium918]

My goal is to use a Drop Down Menu that is populated by

the database. The user is then able to select a name from

the Menu that pulls all of the information for a particular

person from the database. Finally, a table is populated with the information

of the selected person.

[Trium918]

My goal is to use a Drop Down Menu that is populated by

the database. The user is then able to select a name from

the Menu that pulls all of the information for a particular

person from the database. Finally, a table is populated with the information

of the selected person.

[Trium918]

 

 

//Why wouldn't this work?
$query = "select * from books where ".$searchtype." like '%".$searchterm."%'";

//This will display data
$query = "select * from books";

[MadTechie]

whats

$searchtype

$searchterm

set to ?

[AndyB]

More constructively, no it won't produce select * from books it will produce something exactly like your first query with the variables substituted with ... whatever those are/wherever they came from.

[Trium918]

whats

$searchtype

$searchterm

set to ?

 

 

Here it is:

<form action="results.php" method="post">
Choose Search Type:<br>
select name="searchtype">
<option value="author">Author  
<option value="title">Title
<option value="isbn">ISBN
</select>
<br>
Enter Search Term:<br>
<input name="searchterm" type=text>
<br>
<input type=submit value="Search">
</form>

 

AndyB, That is what I think is wrong. The variables get their

value from the form. For some reason it will not display any

data even when I enter something in the textbox.

[MadTechie]

~Sighs~

 

maybe you need to post all your code as that didn't help..

and maybe a DB schemer.