PHP / MySQL: How to reduce code to echo Select results (working code)

Question

I am new to PHP and MySQL and hope you can help me with some tipps or comments on this.

I have a MySQL db with a table "Languages" from which I want to fetch all data and then echo it on my page with some HTML tags around it.

In the example below "ISO" is a language code and "$trans" the selected translation of the language names, e.g. English. The output here is a list of (customized) checkboxes with a checkbox for each language.

So far I have the following which works as intended but since I am new to this I was wondering if I actually need the While loop here or if there is a shorter / better way to get the same - in the end I only need to echo the results of my Select.

My PHP:

// select data from db
$tbl = "Languages";

$conn = new mysqli($servername, $username, $password, $dbname);
$conn->set_charset("utf8");
if($conn->connect_error){
    die("Connection failed: " . $conn->connect_error);
} 
$sql = "SELECT * FROM " . $tbl . " ORDER BY ISO ASC";
$result = $conn->query($sql);  
// create array
while($arrLanguages = $result->fetch_assoc()){
    $languages[] = array("ISO" => $arrLanguages["ISO"], "trans" => $arrLanguages[$lang]);
}
$conn->close();

// echo results with some HTML tags
$i = 1;
foreach($languages as $language){
    echo "<input type='checkbox' class='checkSingle' id='language" . $i . "' name='language' value='" . $language['ISO'] . " - " . $language["trans"] . "' />";
    echo "<label for='language" . $i . "'>" . $language['ISO'] . "</label>";
    $i++;
}

Post your question here codereview.stackexchange.com . stackoverflow is for problems, not for codereviews — TeeDeJee
– TeeDeJee, Commented Jun 16, 2015 at 14:43
don't do select *, then. never suck in fields you're not going to use. just have select ISO, trans, then $languages[] = $arrLanguages. — Marc B
– Marc B, Commented Jun 16, 2015 at 14:44
@MarcB: Thanks a lot for that - this is great ! Will definitely apply. :) If you post it as an answer then I'll accept this. — keewee279
– keewee279, Commented Jun 16, 2015 at 14:45
@TeeDeJee: Thanks - will try again next time. I tried this a few times before but it took very long to get a response there, if any. — keewee279
– keewee279, Commented Jun 16, 2015 at 14:48

Marc B · Accepted Answer · 2015-06-16 14:48:57Z

2

Never do select * if you don't intend to use all of the fields in the table. It's a waste of bandwidth/cpu cycles to request fields that'll simply be thrown away later.

$sql = "SELECT ISO, trans ...";
... execute ...
while($arrLanguages = ...) {
   $languages[] = $arrLanguages;
}

Since you only selected the two fields you really wanted, $arrLanguages will contain ONLY those fields, and you don't have type in the full-blown array creation. That was done automatically by the fetch_assoc() call.

answered Jun 16, 2015 at 14:48

Marc B

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1 Comment

keewee279 Over a year ago

Thanks again, Marc - will follow your suggestion and accept as soon as I can. :)

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