Array to string conversion and Undefined variable in PHP (When used in HTML)

Question

I have been writing a code snippet of PHP in HTML which iterates an array inside another array (data from mysql). But when I run the code, it gives the following two errors.

Undefined variable Array to string conversion in ...... Following is my code.

$sql = "SELECT * FROM person";
$result = mysqli_query($conn, $sql);
$resultDataSet = array();
if (mysqli_num_rows($result) > 0) {

while ($row = mysqli_fetch_array($result)) {
    array_push($resultDataSet, $row);
}
if ($type == "HTML") {
    $htmlSerialize = "
        <table>
            <tr>
                <th>Name</th>
                <th>Age</th>
                <th>City</th>
            </tr>
            <?php foreach($resultDataSet as $value): ?>
             <tr>
                <th><?php=?".$value['Name']." ?></th>
                <th><?php=?".$value['Age']." ?></th>
                <th><?php=?".$value['City']." ?></th>
            </tr>
            <?php endforeach; ?>
        </table>";


    echo $htmlSerialize;

}

Also following are the errors.

What is the error I have made? How can I solve this?

Edited

Following is the var dump of $resultDataSet

array (size=2)
  0 => 
    array (size=8)
      0 => string '1' (length=1)
      'ID' => string '1' (length=1)
      1 => string 'Akila' (length=5)
      'Name' => string 'Akila' (length=5)
      2 => string '22' (length=2)
      'Age' => string '22' (length=2)
      3 => string 'Mount Lavinia' (length=13)
      'City' => string 'Mount Lavinia' (length=13)
  1 => 
    array (size=8)
      0 => string '2' (length=1)
      'ID' => string '2' (length=1)
      1 => string 'Randil' (length=6)
      'Name' => string 'Randil' (length=6)
      2 => string '23' (length=2)
      'Age' => string '23' (length=2)
      3 => string 'Colombo' (length=7)
      'City' => string 'Colombo' (length=7)

Hard to tell what the query is returning. See the output of var_dump($resultDataSet). It could be the columns names are all lowercase. — Adam Azad
– Adam Azad, Commented Sep 25, 2017 at 16:12
@magnus I am a beginner in PHP. Isn't '$value' the object that will be assigned by the for loop for each iteration? — Kapila Hettiarachchi
– Kapila Hettiarachchi, Commented Sep 25, 2017 at 16:13
1. I don't see you have that code inside any loop. You have one loop in this code, the while-loop that you're closing before you're using $value. 2. The variable name will be what you set it to be. I would recommend spending some time going through some basic PHP-tutorials and reading the manual. — M. Eriksson
– M. Eriksson, Commented Sep 25, 2017 at 16:14

Adam Azad · Accepted Answer · 2017-09-25 16:22:04Z

3

The error reports that value is undefined. PHP does not parse nor compile code wrapped by quotes -- strings. Try this where I split the loop from the HTML output.

if ($type == "HTML") {

   // Start by opening the table and header
   $htmlSerialize = '
        <table>
            <tr>
                <th>Name</th>
                <th>Age</th>
                <th>City</th>
            </tr>';

    // loop over the results and append each row to $htmlSerialize
    foreach($resultDataSet as $value):

        $htmlSerialize .= '
                 <tr>
                    <th>'. $value['Name'] .'</th>
                    <th>'. $value['Age'] .'</th>
                    <th>'. $value['City'] .'</th>
                </tr>';

    endforeach;

    // close the table
    $htmlSerialize .= '</table>';

    // flush results
    echo $htmlSerialize;

}

edited Sep 25, 2017 at 16:22

answered Sep 25, 2017 at 16:20

Adam Azad

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