How do you make sure an array is empty in PHP?

Question

Im writing a page in HTML/PHP that connects to a Marina Database(boats,owners etc...) that takes a boat name chosen from a drop down list and then displays all the service that boat has had done on it.

here is my relevant code...

if(isset($_POST['form1'])){//if there was input data submitted
        $form1 = $_POST['form1'];

        $sql1 = 'select Status from ServiceRequest,MarinaSlip where MarinaSlip.SlipID = ServiceRequest.SlipID and BoatName = "'.$form1.'"';
        $form1 = null;
        $result1 = $conn->query($sql1);

        $test = 0;
            while ($row = mysqli_fetch_array($result1, MYSQLI_ASSOC)) {

                    $values1[] = array(
                    'Status' => $row['Status']
                    );
                    $test = 1;

            }

                echo '<p>Service Done:</p><ol>';
                if($test = 1){
                    foreach($values1 as $v1){

                        echo '<li>'.$v1['Status'].'</li>';
                    }
                    echo '</ol>';
                }else{
                    echo 'No service Done';
                }

the issue im having is that some of the descriptions of sevice are simply Open which i do not want displayed as service done, or there is no service completed at all, which throws undefined variable: values1

how would I stop my script from adding Open to the values1 array and display a message that no work has been completed if values1 is empty?

`if($test = 1)` **assigns** `1` to `$test`, and the condition will always evaluate as `true`. `==` compares, with `===` additionally comparing the type. See [**this link**](https://stackoverflow.com/questions/80646/how-do-the-php-equality-double-equals-and-identity-triple-equals-comp) for further information. — Obsidian Age, Apr 19 '18 at 04:08
Possible duplicate of [Reference — What does this symbol mean in PHP?](https://stackoverflow.com/questions/3737139/reference-what-does-this-symbol-mean-in-php) — elixenide, Apr 19 '18 at 05:04
You are wide open to [**SQL injection**](https://www.owasp.org/index.php/SQL_Injection). You need to use prepared statements, rather than concatenating variables into your query. See [How can I prevent SQL injection in PHP?](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php?rq=1). Also, *please* start using more descriptive/meaningful variable and field names. In 6 months, when you need to update this, you're not going to have any idea what's in `$values1`, `form1`, etc. — elixenide, Apr 19 '18 at 05:05

score 1 · Answer 1 · answered Apr 19 '18 at 04:23

1

Try this

    $arr = array();

if (empty($arr)) 
{
     echo'empty array';
}

answered Apr 19 '18 at 04:23

yathavan

183
1
14

Prithwis Chatterjee · Answer 2 · 2018-04-19T04:49:36.193

1

To make sure an array is empty you can use count() and empty() both. but count() is slightly slower than empty().count() returns the number of element present in an array.

$arr=array();
if(count($arr)==0){
//your code here
}

edited Apr 19 '18 at 04:49

answered Apr 19 '18 at 04:41

Prithwis Chatterjee

56
4

score 1 · Answer 3 · answered Apr 19 '18 at 04:44

We often use empty($array_name) to check whether it is empty or not

<?php
if(!empty($array_name))
{
    //not empty
}
else
{
    //empty
}

there is also another way we can double sure about is using count() function

 if(count($array_name) > 0)
 {
    //not empty
 }
 else
 {
    //empty
 }

 ?>

Sameera Prasad Jayasinghe · Answer 4 · 2018-04-19T04:52:45.783

1

try this

if(isset($array_name) && !empty($array_name))
{
//not empty
}

edited Apr 19 '18 at 04:52

answered Apr 19 '18 at 04:47

Sameera Prasad Jayasinghe

1,690
1
13
15

score 0 · Answer 5 · answered Apr 19 '18 at 04:09

0

You can try this- if (empty($somelist)) { // list is empty. }

answered Apr 19 '18 at 04:09

Saif Ali

527
5
9
22

score 0 · Answer 6 · answered Apr 19 '18 at 04:12

0

I often use empty（$arr) to do it.

answered Apr 19 '18 at 04:12

ららら

1
1

score 0 · Answer 7 · answered Apr 19 '18 at 04:40

0

Try this instead:

if (!$values1) {
    echo "No work has been completed";
} else {
    //Do staffs here
}

answered Apr 19 '18 at 04:40

Senura Dissanayake · Answer 8 · 2018-04-21T06:53:18.650

0

All you have to do is get the boolean value of empty($array). It will return false if the array is empty.

You could use empty($varName) for multiple uses. For more reference : http://php.net/manual/en/function.empty.php

edited Apr 21 '18 at 06:53

answered Apr 19 '18 at 05:20

Senura Dissanayake

654
1
9
29

score 0 · Answer 9 · answered Apr 19 '18 at 05:37

I think what you need is to check if $values1 exists so try using isset() to do that and there is no need to use the $test var:

if(isset($values1))
     foreach($values1 as $v1){
         echo '<li>'.$v1['Status'].'</li>';
     }

Or try to define $values1 before the while:

    $values1 = array();

then check if it's not empty:

     if($values1 != '')
         foreach($values1 as $v1){
             echo '<li>'.$v1['Status'].'</li>';
         }

How do you make sure an array is empty in PHP?

9 Answers9