Php echo variables from a function outside the function when calling it

Question

I have this php function which looks up rows from a database table

if (!function_exists("ContactLookup")) {
    function ContactLookup($column, $clause, $value, $limit1=0, $limit2=1)
    {
        global $conn;
        $ContactLookup_Sql= "select * from contacts where $column $clause $value LIMIT $limit1, $limit2";
        $ContactLookup_Rs = mysql_query($ContactLookup_Sql,$conn);
        while ($ContactLookup_Result = mysql_fetch_array($ContactLookup_Rs)) {
            $foremame =$ContactLookup_Result["forename"];
            $surname = $ContactLookup_Result["surname"];
        }
    }
}

This just echoes the results but if I wanted to put the results into a select element how would I do this

Within the while loop would I create the variables then echo the variables when I call the function? Or is there another way around this?

Like:

<?php ContactLookup("company_sequence", "=", $result["contact"],"0","10"); ?>
><select name="contactsequence" id="contactsequence">
    <option value=""><?php echo $forename; ?></option>
        </select

Functions typically should only return or set data, not echo it. Did you try the code you showed as an example? What is the problem? — Jessica
– Jessica, Commented Dec 31, 2013 at 0:07
From my above example I just get an empty select box. Check my edit, I created them as variables and not echoed them — user2135867
– user2135867, Commented Dec 31, 2013 at 0:08
You need to set and return an array, because from how you're calling that function, it could return up to 10 name pairs. — JAL
– JAL, Commented Dec 31, 2013 at 0:29

Jessica · Accepted Answer · 2013-12-31 00:13:00Z

1

Make your function return an array

function ContactLookup ($column, $clause, $value, $limit1=0, $limit2=1) {
    global $conn;
    $ContactLookup_Sql= "select * from contacts where $column $clause $value LIMIT $limit1, $limit2";
    $ContactLookup_Rs = mysql_query($ContactLookup_Sql,$conn);

    $results = [];
    while($ContactLookup_Result = mysql_fetch_array($ContactLookup_Rs))
    {
        $results[] = [
           'forename'=> $ContactLookup_Result["forename"], 
           'surname'=> $ContactLookup_Result["surname"]
        ];
    }
    return $results;
}

Then you can loop through it:

<?php
$names = ContactLookup("company_sequence", "=", $result["contact"],"0","10"); 
echo '<select name="contactsequence" id="contactsequence">';
foreach($names AS $name){
   echo '<option value="">'.$name['forename'].'</option>';
}
echo '</select>';

answered Dec 31, 2013 at 0:13

Jessica

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3 Comments

user2135867 Over a year ago

This is also just returning one row

Jessica Over a year ago

Then your query only returns one row. Print out the query and run it in MySQL manually, what do you get?

Jessica Over a year ago

Try a var_dump on $results do you see more than one result?

JAL · Accepted Answer · 2013-12-31 01:06:14Z

1

Change the function to return the results as a variable, rather than echo them.

function ContactLookup ($column, $clause, $value, $limit1=0, $limit2=1) {
    global $conn;
    $ContactLookup_Sql= "select * from contacts where $column $clause $value LIMIT $limit1, $limit2";
    $ContactLookup_Rs = mysql_query($ContactLookup_Sql,$conn);
    $results=array();
    while($ContactLookup_Result = mysql_fetch_array($ContactLookup_Rs))
    {
        $results[] = array('forename'=>$ContactLookup_Result["forename"],'surname'=>$ContactLookup_Result["surname"];
    }
    return $results;
}

Then in your display loop,

<?php $contacts=ContactLookup("company_sequence", "=", $result["contact"],"0","10"); ?>
<select name="contactsequence" id="contactsequence">
    foreach($contacts as $k=>$contact){?> 
       <option value="<?php echo $k;?>"><?php echo $contact['forename']; ?></option>
        <?php 
        }
   </select>

edited Dec 31, 2013 at 1:06

answered Dec 31, 2013 at 0:13

JAL

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1 Comment

user2135867 Over a year ago

This is just returning one row from the database in my select element

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