How do I embed a .wav file in a php if statment?

Question

I am a noob to php and coding in general so pardon my ignorance. I am trying to play a wav file from a php if statement. When I echo html for the sound I get this error.

PHP Parse error: syntax error, unexpected '" type="' (T_CONSTANT_ENCAPSED_STRING), expecting ',' or ';' in /var/www/html/colums.php on line 67

line 67 being

        echo "<audio autoplay><source src="sounds/beep.wav" type="audio/wav" 
        autoplay></audio>";

Here is my code:

    <?php
    $result = $db->query("SELECT * FROM my_table WHERE sound=1");
    while($row = $result->fetch_assoc()){
    echo $row['sound'];
    }
    $sound=$row['sound'];
    if ($sound = "1"){
            echo "<audio autoplay><source src="/sounds/beep.wav" 
    type="audio/wav" autoplay></audio>";

    }
    ?>

Thanks for your help

score 1 · Accepted Answer · answered May 27 '18 at 05:12

1

the problem within you code lies inside of the quotes. in lots of languages(php included) you can't embed quotes inside of quotes. to reslove your problem you need to add a backslash before every quote inside the quotes that make up the echo. example: echo "";

answered May 27 '18 at 05:12

Nick

69
5

also you do need to fix what nickzor said above and you need to do the same for the while statement because you are using a conditional statement. – Nick May 27 '18 at 05:13
the while is fine – Shadow May 27 '18 at 05:17
I am not sure why the while would work fine, but if it works for you, that's fine. Did my quote suggestion help? – Nick May 27 '18 at 05:21
If you do not know, then I would suggest you check it out. You could start with the sample code in the php manual: http://php.net/manual/en/mysqli-result.fetch-assoc.php – Shadow May 27 '18 at 05:26
You are very much wrong about this. – Shadow May 27 '18 at 05:37
Thanks so much. That was the problem. All working now. – Barrett K May 27 '18 at 06:11

score 0 · Answer 2 · answered May 27 '18 at 05:00

0

The problem from what I can see at a quick glance is in your if statement.

 if ($sound = "1"){

In php a single "=" is assignment of value. "==" and "===" is for comparing the value. So, right now you are assigning the value of 1 to $sound technically. To fix the problem you should just need to replace it with double equals.

if ($sound == "1"){

Also refer to this question for more information.

answered May 27 '18 at 05:00

nickzor

84
7

Thanks for the reply and helpful tip. The problem was more related to Nick's suggestion. But I needed to fix the == too. Thanks for the link. – Barrett K May 27 '18 at 06:14

score 0 · Answer 3 · answered May 27 '18 at 06:13

Found several mistakes in your code ,

You cannot assign $row['sound'] out side the loop. either it should be $row[0]['sound'] or $sound variable should be inside the while loop.
if($sound = 1) will not return error , but logically it's incorrect , it should be ($sound == 1)
you cannot use " inside a string if the string is quoted by " , ( same for ' ). you need to add slashes like \" or \'

Suppose below code will works...

        <?php
    $result = $db->query("SELECT * FROM my_table WHERE sound=1");
    while($row = $result->fetch_assoc()){
    echo $row['sound'];
    $sound=$row['sound'];
    if ($sound == 1){
            echo "<audio autoplay><source src=\"/sounds/beep.wav\" type=\"audio/wav\" autoplay></audio>";
}}
    ?>

How do I embed a .wav file in a php if statment?

3 Answers3