PHP parsing error help (working with MySQL)

Question

It says the error is in the while loop.

    <?php

    $my_connection = mysql_connect('localhost','root','');

    if(!$my_connection){
            die('Could not connect: ' . mysql_error());
    }

    echo 'Connected successfully' . '<br><br>';

    $my_database = mysql_select_db('jihoon');

    if(!$my_database){
        die('Could not find database: ' . mysql_error());
    }

    $query='SELECT * FROM comments";

    $result=mysql_query($query);

    $my_rows;

    while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
        $my_rows = $my_rows . "id : {$row['id']}";
    }

    mysql_close($my_connection);

    ?>

did you see this error: $query='SELECT * FROM comments"; you need to use two of the same brackets quotes. $query="SELECT * FROM comments"; — Tim
– Tim, Commented Jul 24, 2011 at 16:49
hopefully you are using a password for you db and just filled in custom values for post...:) — Tim Specht
– Tim Specht, Commented Jul 24, 2011 at 16:53

Pascal MARTIN · Accepted Answer · 2011-07-24 16:50:47Z

3

If you take a look at the syntax-coloring of your question, here on StackOverflow, you'll notice that it starts getting broken at this line :

$query='SELECT * FROM comments";

You must use the same kind of quote at the beginning and end of your string.

So, you should use either :

$query='SELECT * FROM comments';

or :

$query="SELECT * FROM comments";

But not a bit of both ;-)

answered Jul 24, 2011 at 16:50

Pascal MARTIN

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Dogbert · Accepted Answer · 2011-07-24 16:50:13Z

1

$query='SELECT * FROM comments";

should be

$query="SELECT * FROM comments";

answered Jul 24, 2011 at 16:50

Dogbert

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JK. · Accepted Answer · 2011-07-24 16:50:26Z

1

$query = 'SELECT * FROM comments";

Change to:

$query = "SELECT * FROM comments";

Different quotation marks.

answered Jul 24, 2011 at 16:50

JK.

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devtut · Accepted Answer · 2011-07-24 16:51:26Z

1

You are using a double quote in end $query='SELECT * FROM comments";

answered Jul 24, 2011 at 16:51

devtut

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