php if else - multiple variables not working

Question

I am very new to php and was wondering if somebody would like to weigh in on this one. I'm almost there, but after 3 days I've almost give up :(. I have a table that I that has email address, names and phone #'s. If the person does not have an email address, I would just like the code to print out the name and address with the email link. Sounds simple until I tried to do it :)

Here is my almost working code: (I have a connection already - I works fine as long as I take away trying to eliminate removing the email link part.) Thanks in advance for your time.

    <?php
    $data = mysql_query("SELECT first,last,email,phone FROM church_staff WHERE   
    display='yes' AND pull_justice='yes' ORDER BY last ASC")
    or die(mysql_error());  
    while($info = mysql_fetch_array( $data ))
    Print " <tr><td class=\"low\"> ";
    // Evaluates to true because $email is empty
    if (empty($email)) {
    Print " ".$info['first']." ".$info['last']."  ";
    }
    // Evaluates as true because $email is set
    if (isset($email)) {
    }
    Print " <img src=\"images/email.gif\" alt=\"email\">
    <a  href=\"mailto:".$info['email']." >".$info['first']." ".$info['last']."</a> ";
    Print " </td><td class=\"low\">".$info['phone']."</td></tr> ";
    ?>

Can you clarify: "I would just like the code to print out the name and address with the email link." — deed02392
– deed02392, Commented Apr 3, 2013 at 14:23

Marcassin · Accepted Answer · 2013-04-03 14:28:52Z

1

You should use $info['email'] instead of $email, because this variable $email is never set.

You Forgot to add bracketS after your while condition :

    while ($info = mysql_fetch_array ($data))
{
 // process on $info
}

so in your code you print " <tr><td class=\"low\"> "; n times (n the number of returned rows by your query)

answered Apr 3, 2013 at 14:28

Marcassin

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Comments

kainaw · Accepted Answer · 2013-04-03 14:29:16Z

1

First, $email is not set. It is $info['email']. You will always have $info['email'], but the value may be null or "". So, all you need to do is check if($info['email'] == "") to see if it is not set. Technically, I would check if(trim($info['email']) == "") because I do not trust data entry people. They may stick blank spaces or tabs in there.

Now, you want an if else:

if(trim($info['email')=="")
{
    //do what you want when there is no email
}
else
{
    //do what you want there this an email
}

answered Apr 3, 2013 at 14:29

kainaw

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Comments

FernCoder · Accepted Answer · 2013-04-03 14:35:08Z

0

I'm not really sure of what you are trying to do it look like I missed something. But first of all your code would be better like this :

<?php
$data = mysql_query("SELECT first,last,email,phone FROM church_staff WHERE   
display='yes' AND pull_justice='yes' ORDER BY last ASC")
or die(mysql_error());  
while($info = mysql_fetch_array( $data ))
{
    Print " <tr><td class=\"low\"> ";
    // Evaluates to true because $email is empty
    // Evaluates as true because $email is set
    if (isset($email)) {        
        if (empty($email)) {
            Print " ".$info['first']." ".$info['last']."  ";
        }
    }
    Print " <img src=\"images/email.gif\" alt=\"email\">
    <a  href=\"mailto:".$info['email']." >".$info['first']." ".$info['last']."</a> ";
    Print " </td><td class=\"low\">".$info['phone']."</td></tr> ";
}
?>

Indeed, While loops will use the first line after the declaration if no scope is define. Plus, you better check if something is set before checking if it's empty or not.

Hope I answered your question.

answered Apr 3, 2013 at 14:35

FernCoder

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2 Comments

nibra Over a year ago

isset is part of, what empty tests. Your correction always prints an email link, regardless the email address' existence.

Jay Jones Over a year ago

I am just printing out a list of names, phone and numbers and email address'. The email address' are a mailto link of the first name and last name. However, not everybody has an email address. So I would like to NOT show and email link if there is NO email address in the database. I would only like to print out the first and last name. I have tried both examples above - thanks so much for your efforts :) but they do not work for me - close though...

nibra · Accepted Answer · 2013-04-03 14:51:05Z

Your while loop contains only one print statement; everything else is outside of it, because you forgot the curly braces. Next, the closing curly brace for the second if's body comes too early. After correcting that, your loop looks like:

while ($info = mysql_fetch_array($data)) {
    print " <tr><td class=\"low\"> ";
    if (empty($email)) {
        print " ".$info['first']." ".$info['last']."  ";
    }
    if (isset($email)) {
        print " <img src=\"images/email.gif\" alt=\"email\"><a  href=\"mailto:".$info['email']." >".$info['first']." ".$info['last']."</a> ";
    }
    print " </td><td class=\"low\">".$info['phone']."</td></tr> ";
}

If you had turned *error_reporting* to E_ALL, like you always should, when developing, you'd see, that $email is not defined. Obviously, it should be $info['email'] instead:

while ($info = mysql_fetch_array($data)) {
    print " <tr><td class=\"low\"> ";
    if (empty($info['email'])) {
        print " ".$info['first']." ".$info['last']."  ";
    }
    if (isset($info['email'])) {
        print " <img src=\"images/email.gif\" alt=\"email\"><a  href=\"mailto:".$info['email']." >".$info['first']." ".$info['last']."</a> ";
    }
    print " </td><td class=\"low\">".$info['phone']."</td></tr> ";
}

Now, a variable in PHP can be both set and empty, so it is possible that the name is printed twice. to avoid that, turn the second if into an else branch.

while ($info = mysql_fetch_array($data)) {
    print " <tr><td class=\"low\"> ";
    if (empty($info['email'])) {
        print " ".$info['first']." ".$info['last']."  ";
    } else {
        print " <img src=\"images/email.gif\" alt=\"email\"><a  href=\"mailto:".$info['email']." >".$info['first']." ".$info['last']."</a> ";
    }
    print " </td><td class=\"low\">".$info['phone']."</td></tr> ";
}

To keep things clear, you should always avoid to mix code and layout as much as possible. The next step thus is to extract the HTML templates and add the image using CSS to a.mail.

$template_container = '<tr><td class="low">%s</td><td class="low">%s</td></tr>';
$template_mail      = '<a href="mailto:%s">%s</a>';

while ($info = mysql_fetch_array($data)) {
    $name = $info['first']." ".$info['last'];
    if (!empty($info['email'])) {
        $name = sprintf($template_mail, $info['email'], $name);
    }
    printf($template_container, $name, $info['phone']);
}

As you can see, the code is now much easier to read.

With a bit of editing this one worked..while ($info = mysql_fetch_array($data)) { print " <tr><td class=\"low\"> "; if (empty($info['email'])) { print " ".$info['first']." ".$info['last']." "; } else { print " <img src=\"images/email.gif\" alt=\"email\"><a href=\"mailto:".$info['email']." >".$info['first']." ".$info['last']."</a> "; } print " </td><td class=\"low\">".$info['phone']."</td></tr> "; } Thank you so much for your help - I'm learning, and you helped. I appreciate your time. I really do :)

MaNKuR · Accepted Answer · 2013-04-03 15:02:32Z

0

** See the code rectified the error **

<?php
$data = mysql_query("SELECT first,last,email,phone FROM church_staff WHERE   
display='yes' AND pull_justice='yes' ORDER BY last ASC")
or die(mysql_error());  
while($info = mysql_fetch_array( $data ))
{
    // Confirm if $data is having 1d array of required index and value
    Print " <tr><td class=\"low\"> ";
    // Evaluates to true because $email is empty
    if (empty($data['email'])) {
        Print " ".$info['first']." ".$info['last']."  ";
    }
    else
    {
        Print " <img src=\"images/email.gif\" alt=\"email\">
        <a  href=\"mailto:".$info['email']." >".$info['first']." ".$info['last']."</a> ";
        Print " </td><td class=\"low\">".$info['phone']."</td></tr> ";
    }
}
?>

Better learn php's syntext in delve ... i.e isset() is a subset of empty.

edited Apr 3, 2013 at 15:02

answered Apr 3, 2013 at 14:34

MaNKuR

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4 Comments

nibra Over a year ago

$data['email'] is not defined.

MaNKuR Over a year ago

but don't you think @nibra we should also test the number of field return by the query and have if condition just before the while like if(mysql_num_rows($data) > 0) { Otherwise it will throw an error if $data return empty result. mysql_fetch_array manual

nibra Over a year ago

no, while is testing that already. $data contains a valid resource at that point. If no (more) records are present, null is returned, which makes the while-condition evaluate to false.

MaNKuR Over a year ago

I have already experienced with issues [bytes.com/topic/mysql/answers/… and its a best practice to handle exception at the beginning.

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