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elseif conditional

Daveyboy

By Daveyboy
in PHP Coding Help

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Daveyboy Newbie

Daveyboy

Posted
Posted
Can't figure this out. Without the elseif's it works fine, but with the elseif's it doesn't do anything, no errors.

[code]
<html>
<body>

<form action="databasetest.php" method="post">
<fieldset>
<p><b>First Name: </b><input type="text" name="firstname" size="60" maxlength="60" value= ><br /></p>

<p><b>Last Name:</b> <input type="text" name="lastname" size="60" maxlength="60" value= ><br />

<p><b>Age:</b> <input type="text" name="age" size="60" maxlength="60" value=  ><br />

</fieldset>
<input type="hidden" name="submitted" value="TRUE" />
<div align="center"><input type="submit" name="sumbit" value="Submit" /></div>
</form>

</body>
</html>

<?php
error_reporting(E_ALL);
require_once ("./mysql_connect_databasetest.php");
if (isset($_POST['submitted'])){
}elseif (isset($_POST['firstname']) && (is_numeric(($_POST['firstname'])))) {
echo 'please type in valid first name';
}elseif(isset($_POST['lastname']) && (is_numeric(($_POST['lastname'])))) {
echo 'please type in a valid last name';

$firstname= $_POST['firstname'];
$lastname= $_POST['lastname'];
$age= $_POST['age'];
$query = "INSERT INTO info (firstname,lastname,age) VALUES ('$firstname','$lastname','$age')";
mysql_query($query);
    echo "thank you for your submission";
}else{
echo mysql_error();
}
print_r($query);
?>
[/code]
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dbo Member

dbo

Posted
Posted
[quote author=Daveyboy link=topic=120439.msg493986#msg493986 date=1167526627]
not according to php.net

http://ca.php.net/manual/en/control-structures.elseif.php
[/quote]

Very interesting. Never wrote it that way. Did you figure it out?
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dcro2 Newbie

dcro2

Posted
Posted
I don't think we understand most of this.
I don't think this is what you meant to do, though:

[b]If the form was submitted then:[/b]
    Nothing;
[b]Else If the first name is set and it is numeric then:[/b]
    Say 'please type in valid first name';
[b]Else If the last name is set and it is numeric then:[/b]
    Say 'please type in a valid last name';
    $firstname= $_POST['firstname'];
$lastname= $_POST['lastname'];
$age= $_POST['age'];
$query = "INSERT INTO info (firstname,lastname,age) VALUES ('$firstname','$lastname','$age')";
mysql_query($query);
echo "thank you for your submission";
[b]If none of these things are true then: (What things?)[/b]
    echo mysql_error();
[b]End If[/b]
print_r($query);

You see, I don't think you mean to put in the first [b]elseif[/b] because it should have to go if the form was submitted.
Let me suggest some code, maybe it's right:

[code]<?php
error_reporting(E_ALL);

require_once ("./mysql_connect_databasetest.php");

if ( isset($_POST['submitted']) )
{
if ( isset($_POST['firstname']) && is_numeric($_POST['firstname']) )
{
echo 'Please type in valid first name.';
exit; //Only using exit as example
}
if ( isset($_POST['lastname']) && is_numeric($_POST['lastname']) )
{
echo 'Please type in a valid last name.';
exit; //Only using exit as example
}

//If none of those things are true...

$firstname= $_POST['firstname'];
$lastname= $_POST['lastname'];
$age = $_POST['age'];
$query = "INSERT INTO info (firstname,lastname,age) VALUES ('$firstname','$lastname','$age')";

if( mysql_query($query) )
echo "Thank you for your submission.";
}else{
echo mysql_error();
}
}
print_r($query);
?>[/code]
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Adika Newbie

Adika

Posted
Posted
Here is the solution:

[code]
<?php
error_reporting(E_ALL);

require_once ("./mysql_connect_databasetest.php");

if ( isset($_POST['submitted']) )
{
if ( isset($_POST['firstname']) && is_numeric($_POST['firstname']) )
{
echo 'Please type in valid first name.';
exit();
}
if ( isset($_POST['lastname']) && is_numeric($_POST['lastname']) )
{
echo 'Please type in a valid last name.';
exit();
}
if(!isset($_POST['lastname']) || !isset($_POST['firstname']) {
                      exit();
          } else {
//If none of those things are true...

$firstname= $_POST['firstname'];
$lastname= $_POST['lastname'];
$age = $_POST['age'];
$query = "INSERT INTO info (firstname,lastname,age) VALUES ('$firstname','$lastname','$age')";

if( mysql_query($query) ) {
echo "Thank you for your submission.";
} else {
echo mysql_error();
}
}
}
print_r($query);
?>
[/code]
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Daveyboy Newbie

Daveyboy

Posted
  • Author
Posted
No it still does not work, the logic should make sense. I wonder if it has something to do with the html form?

[code]<html>
<body>

<form action="databasetest.php" method="post">
<fieldset>
<p><b>First Name: </b><input type="text" name="firstname" size="60" maxlength="60" value= ><br /></p>

<p><b>Last Name:</b> <input type="text" name="lastname" size="60" maxlength="60" value= ><br />

<p><b>Age:</b> <input type="text" name="age" size="60" maxlength="60" value=  ><br />

</fieldset>
<input type="hidden" name="submitted" value="TRUE" />
<div align="center"><input type="submit" name="sumbit" value="Submit" /></div>
</form>

<?php
error_reporting(E_ALL);

require_once ("./mysql_connect_databasetest.php");

if ( isset($_POST['submitted']) )
{
if ( isset($_POST['firstname']) && is_numeric($_POST['firstname']) )
{
echo 'Please type in valid first name.';
exit();
}
if ( isset($_POST['lastname']) && is_numeric($_POST['lastname']) )
{
echo 'Please type in a valid last name.';
exit();
}
if (!isset($_POST['lastname']) || !isset($_POST['firstname']) ) {
                      exit();
          } else {
//If none of those things are true...

$firstname= $_POST['firstname'];
$lastname= $_POST['lastname'];
$age = $_POST['age'];
$query = "INSERT INTO info (firstname,lastname,age) VALUES ('$firstname','$lastname','$age')";

if( mysql_query($query) ) {
echo "Thank you for your submission.";
} else {
echo mysql_error();
}
}
}
print_r($query);
?>

</body>
</html>[/code]
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matto Newbie

matto

Posted
Posted
unless you are setting some default values there is no need to have the follow:

[code]
<input type="text" name="firstname" size="60" maxlength="60" value= >
[/code]
This should probably be:
[code]
<input type="text" name="firstname" size="60" maxlength="60">
[/code]

looking a little closer at your code something looks a bit odd

[code]
if ( isset($_POST['firstname']) && is_numeric($_POST['firstname']) )
{
echo 'Please type in valid first name.';
exit();
}
[/code]

The above is suggesting if firstname is provided and firstname is numeric then throw an error and exit - it's doesn't suggest this is to be a mandatory field. If you are wanting this to be a mandatory field, this may be better.

[code]
if (!isset($_POST['firstname']) || is_numeric($_POST['firstname']) )
{
echo 'Please type in valid first name.';
exit();
}
[/code]

:)
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matto Newbie

matto

Posted
Posted
Give this a go...

[code]
<html>
<body>
<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="post">
<fieldset>
<p><b>First Name: </b><input type="text" name="firstname" size="60" maxlength="60"><br /></p>
<p><b>Last Name:</b> <input type="text" name="lastname" size="60" maxlength="60"><br />
<p><b>Age:</b> <input type="text" name="age" size="60" maxlength="60" value=  ><br />
</fieldset>
<div align="center"><input type="submit" name="submit" value="Submit" /></div>
</form>
<?php
error_reporting(E_ALL);
require_once("./mysql_connect_databasetest.php");
if (isset($_POST['submit']) ) { 
  if (empty($_POST['firstname']) || is_numeric($_POST['firstname']) ) {
    echo 'Please type in valid first name.';
    exit;
  }
  if (empty($_POST['lastname']) || is_numeric($_POST['lastname']) ) {
    echo 'Please type in a valid last name.';
    exit;
  }
  $query = "INSERT INTO info (firstname,lastname,age) VALUES ('" . $_POST['firstname'] . "','" . $_POST['lastname'] . "','" . $_POST['age'] . "')";
 
  if( mysql_query($query) ) {
    echo "Thank you for your submission."; 
  } else {
    echo mysql_error();
  }
}
?>
</body>
</html>
[/code]

;)
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