still struggling - please help

Question

phpNewbie 0 Newbie Poster

15 Years Ago

i'm still struggling with php - its soooo confusing to me

this is the code i have now - is it in the wrong order? or whats missing please to make it work - i'm trying to get a result if 'both' genders are selected and if 'any' origin is selected

thanks in advance for any help

<html>
	<head>
	</head>
	<body>
		<?php
			// MySQL Connection Information
			$server = "localhost";
			$username = "sharlene";
			$password = "";
			$dbname = "baby_names";

			// MySQL Connect String
			$connection = mysql_connect($server,$username,$password) or die("Can't Connect to Mysql Server: 

".mysql_error());

			// Select the database
			mysql_select_db($dbname) or die("Can't connect to database: ".mysql_error());

			$gender = mysql_real_escape_string($_POST['gender']);
			$meaning = mysql_real_escape_string($_POST['meaning']);
			$name = mysql_real_escape_string($_POST['name']);
			$origin = mysql_real_escape_string($_POST['origin']);


  
$sql = "SELECT * FROM names WHERE gender = '" . $gender . "' and name LIKE '" . $name . "%' and 
meaning LIKE '%" . $meaning . "%' and origin = '" . $origin . "'";
  

if ($gender == 'both') { // no specific gender
  if ($origin == 'any') { // no specific origin
    $sql = 'select * from names';
  else { // an origin was specified
    $sql = "select * from names where origin = '".$origin."'"; 
  }

else { // a gender was specified
  
  if ($origin == 'any') { // no specific origin
    $sql = "select * from names where gender ='".$gender."'";
  else { // an origin was also specified
    $sql = "select * from names where origin = '".$origin."' and  gender ='".$gender."'"; 
}

			
			//execute SQL query and get result
			$sql_result = mysql_query($sql, $connection) or die(mysql_error());
			
		?>
			<table border="0" align="center">
				<tr>
					<th style="background: #ffd" >Name</th>
					<th style="background: #efe">Gender</th>
					<th style="background: #fef">Origin</th>
					<th style="background: #e3e4fa">Meaning</th>
				</tr>
				
		<?php
			// Loop through the data set and extract each row into it's own variable set
			while ($row = mysql_fetch_array($sql_result)) 
			{
				//extract($row);
			?>
				<tr>
					<td style="background: #ffd"><?php echo $row['name'];?></td>
					<td style="background: #efe"><?php echo $row['gender'];?></td>
					<td style="background: #fef"><?php echo $row['origin'];?></td>
					<td style="background: #e3e4fa"><?php echo $row['meaning'];?></td>
				</tr>
			<?php
			}
			
		?>
	</body>
</html>

my form is:

<!--html page-->
<html>
	<head>
		<title>Baby Names Database</title>
	</head>
	<body>
		<h1>Baby Names</h1>
		<h3>Please fill in the blanks below, and We'll return baby names for you to browse.</h3>

		
			<table border = "1" align = "center">
				<form method = "post" action = "babyNames1.php">
				<tr bgcolor="#cae1ff">
					<td align = "left">
						Gender:
					</td>
					<td>
						<input type = "radio" name = "gender" value = "male"/>Boys Names
						<input type = "radio" name = "gender" value = "female"/>Girls Names
						<input type = "radio" name = "gender" value = ""/>Both
					</td>
				</tr>
				<tr bgcolor="#ffe1ff">
					<td align = "left">
						Meaning:
					</td>
					<td>
						<input type = "text" name = "meaning" value = ""/>
					</td>
				</tr>
				<tr bgcolor="#ffe1ff">
					<td align = "left">
						Name Begins With:
					</td>
					<td>
						<input type = "text" name = "name" value = ""/>
					</td>
				</tr>
				<tr bgcolor="#cae1ff">
					<td align = "left">
						Origin
					</td>
					<td>
					
						<select name = "origin" id = "origin">
							<option value = "any">Any</option>
							<option value = "african">African</option>
							<option value = "anglo">Anglo</option>
							<option value = "arabian">Arabian</option>
							<option value = "arabic">Arabic</option>
							<option value = "aramaic">Aramaic</option>
							<option value = "armenian">Armenian</option>
							<option value = "arthurian">Arthurian</option>
							<option value = "basque">Basque</option>
							<option value = "celtic">Celtic</option>
							<option value = "chamoru">Chamoru</option>
						</select> 
					</td>
				</tr>
			
			</table><p /><center>
<input type = "submit" value = "Show me Names"/>
</center>	
</form>
</body>
</html>

php

6 Contributors
14 Replies
141 Views
2 Weeks Discussion Span
Latest Post 15 Years Ago Latest Post by phpNewbie

All 14 Replies

Murtan 317 Practically a Master Poster

15 Years Ago

I think the problem with the 'both' option is that when you select 'both' on the form, gender is set to '' and not to 'both'.

cwarn23 387 Occupation: Genius

15 Years Ago

BTW, just to give you an idea of where Murtan is talking about (not very obvious) it is the following line of the second file where value=

<input type = "radio" name = "gender" value = ""/>Both

As you can see there is no value assigned to that field and that is why it will appear as empty in the database.

vinothkumarc 9 Light Poster

15 Years Ago

<html>
	<head>
	</head>
	<body>
		<?php
			// MySQL Connection Information
			$server = "localhost";
			$username = "sharlene";
			$password = "";
			$dbname = "baby_names";

			// MySQL Connect String
			$connection = mysql_connect($server,$username,$password) or die("Can't Connect to Mysql Server: 

".mysql_error());

			// Select the database
			mysql_select_db($dbname) or die("Can't connect to database: ".mysql_error());

			$gender = mysql_real_escape_string($_POST['gender']);
			$meaning = mysql_real_escape_string($_POST['meaning']);
			$name = mysql_real_escape_string($_POST['name']);
			$origin = mysql_real_escape_string($_POST['origin']);
      if($gender) {
        $whereArr[] = "gender = '" . $gender . "' ";
      }
      if($name) {
        $whereArr[] = "name LIKE '" . $name . "%'";
      }
      if($meaning) {
        $whereArr[] = "meaning LIKE '%" . $meaning . "%'";
      }
      if($origin) {
        $whereArr[] = "origin = '" . $origin . "' ";
      }
      if(count($whereArr)) {
        $where = @implode(" AND ", $whereArr);
        $where = ' WHERE '.$where;
      }
  
$sql = "SELECT * FROM names $where";
  
/*
if ($gender == 'both') { // no specific gender
  if ($origin == 'any') { // no specific origin
    $sql = 'select * from names';
  else { // an origin was specified
    $sql = "select * from names where origin = '".$origin."'"; 
  }

else { // a gender was specified
  
  if ($origin == 'any') { // no specific origin
    $sql = "select * from names where gender ='".$gender."'";
  else { // an origin was also specified
    $sql = "select * from names where origin = '".$origin."' and  gender ='".$gender."'"; 
}
/**/
			
			//execute SQL query and get result
			$sql_result = mysql_query($sql, $connection) or die(mysql_error());
			
		?>
			<table border="0" align="center">
				<tr>
					<th style="background: #ffd" >Name</th>
					<th style="background: #efe">Gender</th>
					<th style="background: #fef">Origin</th>
					<th style="background: #e3e4fa">Meaning</th>
				</tr>
				
		<?php
			// Loop through the data set and extract each row into it's own variable set
			while ($row = mysql_fetch_array($sql_result)) 
			{
				//extract($row);
			?>
				<tr>
					<td style="background: #ffd"><?php echo $row['name'];?></td>
					<td style="background: #efe"><?php echo $row['gender'];?></td>
					<td style="background: #fef"><?php echo $row['origin'];?></td>
					<td style="background: #e3e4fa"><?php echo $row['meaning'];?></td>
				</tr>
			<?php
			}
			
		?>
	</body>
</html>

phpNewbie commented: works wonderfully! +1

nav33n 472 Purple hazed!

15 Years Ago

start quote:

<html>
    <head>
    </head>
    <body>
        <?php
            // MySQL Connection Information
            $server = "localhost";
            $username = "sharlene";
            $password = "";
            $dbname = "baby_names";

            // MySQL Connect String
            $connection = mysql_connect($server,$username,$password) or die("Can't Connect to Mysql Server: 

".mysql_error());

            // Select the database
            mysql_select_db($dbname) or die("Can't connect to database: ".mysql_error());

            $gender = mysql_real_escape_string($_POST['gender']);
            $meaning = mysql_real_escape_string($_POST['meaning']);
            $name = mysql_real_escape_string($_POST['name']);
            $origin = mysql_real_escape_string($_POST['origin']);
      if($gender) {
        $whereArr[] = "gender = '" . $gender . "' ";
      }
      if($name) {
        $whereArr[] = "name LIKE '" . $name . "%'";
      }
      if($meaning) {
        $whereArr[] = "meaning LIKE '%" . $meaning . "%'";
      }
      if($origin) {
        $whereArr[] = "origin = '" . $origin . "' ";
      }
      if(count($whereArr)) {
        $where = @implode(" AND ", $whereArr);
        $where = ' WHERE '.$where;
      }

$sql = "SELECT * FROM names $where";

/*
if ($gender == 'both') { // no specific gender
  if ($origin == 'any') { // no specific origin
    $sql = 'select * from names';
  else { // an origin was specified
    $sql = "select * from names where origin = '".$origin."'"; 
  }

else { // a gender was specified

  if ($origin == 'any') { // no specific origin
    $sql = "select * from names where gender ='".$gender."'";
  else { // an origin was also specified
    $sql = "select * from names where origin = '".$origin."' and  gender ='".$gender."'"; 
}
/**/

            //execute SQL query and get result
            $sql_result = mysql_query($sql, $connection) or die(mysql_error());

        ?>
            <table border="0" align="center">
                <tr>
                    <th style="background: #ffd" >Name</th>
                    <th style="background: #efe">Gender</th>
                    <th style="background: #fef">Origin</th>
                    <th style="background: #e3e4fa">Meaning</th>
                </tr>

        <?php
            // Loop through the data set and extract each row into it's own variable set
            while ($row = mysql_fetch_array($sql_result)) 
            {
                //extract($row);
            ?>
                <tr>
                    <td style="background: #ffd"><?php echo $row['name'];?></td>
                    <td style="background: #efe"><?php echo $row['gender'];?></td>
                    <td style="background: #fef"><?php echo $row['origin'];?></td>
                    <td style="background: #e3e4fa"><?php echo $row['meaning'];?></td>
                </tr>
            <?php
            }

        ?>
    </body>
</html>

end quote.

Use code tags for FS.

Edited 11 Years Ago by mike_2000_17 because: Fixed formatting

Reply to this topic

Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.

vinothkumarc 9 Light Poster · Answer 1 · 2009-03-03T13:35:07+00:00

hey... y u copied... and pasted without any comments (my code)

phpNewbie 0 Newbie Poster · Answer 2 · 2009-03-03T13:53:18+00:00

hey... y u copied... and pasted without any comments (my code)

umm... i dont know you, have never heard of you and its not the same code... a friend helped me with it... do you have another user name?

Murtan 317 Practically a Master Poster · Answer 3 · 2009-03-03T14:09:02+00:00

I think nav33n (who re-posted your code) was trying to encourage you to use code tags.

Please use code tags when posting code. For more information, click here.

When posting php code, please use php code tags
[code=php] // Your code here

[/code]

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 4 · 2009-03-03T14:09:35+00:00

hey... y u copied... and pasted without any comments (my code)

:icon_rolleyes:

cwarn23 387 Occupation: Genius Team Colleague Featured Poster · Answer 5 · 2009-03-03T14:20:59+00:00

hey... y u copied... and pasted without any comments (my code)

It appears he didn't really copy your code but instead quoted your post.

phpNewbie 0 Newbie Poster · Answer 6 · 2009-03-03T16:34:35+00:00

sorry people - i thought i did use the tags but anyway thank you all for your replies - will play with it tomorrow

nav33n 472 Purple hazed! Team Colleague Featured Poster · Answer 7 · 2009-03-03T16:43:08+00:00

sorry people - i thought i did use the tags but anyway thank you all for your replies - will play with it tomorrow

@phpNewbie, We aren't talking about you. Vinothkumarc didn't use [code] tags. Infact, I appreciate you for using code tags. :)

emiola 0 Junior Poster in Training · Answer 8 · 2009-03-04T01:47:10+00:00

Hello Newbie. You don't need to do so much. Really PHP script is simple to write. all you need is make your code like a lawyer builds a case. Let's see if I can use your code to explain somethings to you but right away, I have an issue with a script too. See you in about 10 hours time.
Feel cool, PHP is friendly.

phpNewbie 0 Newbie Poster · Answer 9 · 2009-03-04T10:02:49+00:00

Hello Newbie. You don't need to do so much. Really PHP script is simple to write. all you need is make your code like a lawyer builds a case. Let's see if I can use your code to explain somethings to you but right away, I have an issue with a script too. See you in about 10 hours time.
Feel cool, PHP is friendly.

hey thanks... i look forward to this too

phpNewbie 0 Newbie Poster · Answer 10 · 2009-03-19T04:42:35+00:00

Hey thanks so much vinothkumarc! You're awesome... sorry it took me so long to get back and thank you for solving.

still struggling - please help

Recommended Answers Collapse Answers

All 14 Replies

Recommended Answers