How can I stop my PHP from spamming the error.log with PHP Notice?

Question

I'm currently working on a project where I need to get some data from 3 different tables out of a database and echo them so I can work with the results in jQuery. I'm sending a GET-request with 3 variables to my PHP. The first one is used to determine which 'command' needs to be executed, the other 2 are used to determine which table and which row needs to be queried.

This is the code I have so far:

} elseif($_GET['command']=='getpage') {
    $mypid = $_GET['id'];
    $mytable = $_GET['table'];

    $link = mysqli_connect($dbserver,$userdb,$passdb,$db_typo) or die(mysqli_error($link));

    if($mytable == 'tableName1'){
        $query = 'SELECT * FROM table1 WHERE uid = "'.$mypid.'"'; //I need 6 elements from this table
    } elseif($mytable == 'tableName2'){
        $query = 'SELECT * FROM table2 WHERE uid = "'.$mypid.'"'; //I need 7 elements from this table
    } elseif($mytable =='tableName3'){
        $query = 'SELECT * FROM table3 WHERE uid = "'.$mypid.'"'; //I need 8 elements from this table
    } else {
        echo 'no such table supported for this command';
    }

    $result = mysqli_query($link, $query) or die(mysqli_error($link));

    $pagecontent = array();
    while($row = mysqli_fetch_assoc($result)){
        $pagecontent[] = array(
            'id' => utf8_encode($row['uid']),
            'name' => utf8_encode($row['name']),
            'text1' => utf8_encode($row['text1']), //not every table has this
            'text2' => utf8_encode($row['text2']),
            'img' => utf8_encode($row['image']),
            'parent' => utf8_encode($row['parent']), //not every table has this
            'sub_parent' => utf8_encode($row['sub_parent']), //not every table has this
            'deleted' => utf8_encode($row['deleted'])
        );
    }

    echo '{"content": '.json_encode($pagecontent).'}';
}

I have over 50 pages which I need to get from the database. So when I would let the jQuery function that sends the GET-request run trough I would end up spamming the error.log with

PHP Notice: Undefined index: text1 in /var/www/my.php on line 171

which I don't want.

Is there another way to fix this 'problem' than just putting the query and while-loop inside the if-statement?

add the column text1 in every table, even if it will be empty. — DocRattie
– DocRattie, Commented May 6, 2015 at 13:18
why wouldn't you want to not have a checking if condition for index testing? — Kevin
– Kevin, Commented May 6, 2015 at 13:19
@DocRattie The tables are created automatically. If I would put them together my project wouldn't work. — Timo
– Timo, Commented May 6, 2015 at 13:19
why dont you check if the field exist in table or not then allow the value .. — lakshman
– lakshman, Commented May 6, 2015 at 13:20

ihsan · Accepted Answer · 2015-05-06 13:23:52Z

3

Add a check whether an array key exists

'text1' => isset($row['text1']) ? utf8_encode($row['text1']) : '',

answered May 6, 2015 at 13:23

ihsan

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Kevin · Accepted Answer · 2015-05-06 13:22:11Z

1

I don't know whats the reason of why don't you want an if to check for indices, but you could add another loop inside the while

and add those variables:

while($row = mysqli_fetch_assoc($result)){
    $temp = array();
    foreach($row as $key => $value) {
        $temp[$key] = utf8_encode($value);
    }
    $pagecontent[] = $temp;
    // $pagecontent[] = array_map('utf8_encode', $row);
}

echo json_encode(array('content' => $pagecontent));

This just simply gets all the contents inside your $row, encodes all values then push it inside your container.

Sidenote: Don't build the JSON string by hand, create the final array structure, then encode in the end.

answered May 6, 2015 at 13:22

Kevin

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