why does foreach skip 2 rows from the query result?

Question

For 10 rows in the query it only returns 8 rows but i get the fields right:

For Query data which returns less than 2 rows I get an error.

//create table to display all data
echo "<table border="1"> ";
echo "<tr>";
$row = mysqli_fetch_assoc($result);
    foreach($row as $key => $value) 
        {
            echo "<th>$key</th>";
        }
echo "</tr>";

while (($row = $result->fetch_assoc()) !== null) 
{
  $output = array();
  $i=1;

  echo "<tr>";

    foreach ($row as $columnName => $columnValue) 
        {
            $output[] = $columnName."=>". $columnValue;
            echo "<td>".$columnValue."</td>";
        }

  echo "</tr>";
}
echo "</table>";

Please post the foreach in its correct context of your fetch loop. I suspect there's something going on like calling mysqli_fetch_row() both inside and outside a loop, causing a row to be skipped since the result resource isn't rewound. — Michael Berkowski
– Michael Berkowski, Commented Nov 26, 2014 at 18:54
here is my query for 1 row of $result. I get no output. $row=mysqli_fetch_assoc($result); if (is_array($row)) { echo "Yes, this is an array<br>"; foreach($row as $key => $value) { echo"$key\n"; } } — Kumar Gavade
– Kumar Gavade, Commented Nov 26, 2014 at 18:58
the "invalid argument supplied for foreach()" part is probably due to not checking for mysqli_fetch_row() result against false which is returned instead of an array if there are no more rows to fetch — Hartmut Holzgraefe
– Hartmut Holzgraefe, Commented Nov 26, 2014 at 18:59
Do you mean to say that it skips the first two columns and only returns 8 columns ? The difference is significant from rows. — Michael Berkowski
– Michael Berkowski, Commented Nov 26, 2014 at 19:01
Adding my modified code in the original question where I am creating a table with 10 rows on data it only returns 8 — Kumar Gavade
– Kumar Gavade, Commented Nov 26, 2014 at 19:03

pradeepcep · Accepted Answer · 2014-11-26 19:32:26Z

2

Edit: Thanks to @Michael Berkowski for his comments on the question.

Here is a modified version of your code that should work:

//create table to display all data
echo "<table border=1 >";
echo "<tr>";
//echo "<th> ## </th>";

$row = $result->fetch_assoc();  // stick to the object-oriented form. It is cleaner.
foreach ($row as $key => $value) 
{
    echo "<th>$key</th>";
}
echo "</tr>";

do
{
    $output = array();
    echo "<tr>";
    foreach ($row as $columnName => $columnValue) 
    {
        $output[$columnName] = $columnValue;  // this is neater.
        echo "<td>" . $columnValue . "</td>";
    }
    echo "</tr>";
} while ($row = $result->fetch_assoc());

echo "</table>";

You can use your first foreach() loop to print the keys and then use a do-while() loop to get your desired output.

Additional reading:

PHP do-while loop

answered Nov 26, 2014 at 19:32

pradeepcep

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Comments

Pupil · Accepted Answer · 2014-11-26 18:54:18Z

-1

You need to use

mysqli_fetch_assoc($result);

answered Nov 26, 2014 at 18:54

Pupil

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3 Comments

Michael Berkowski Over a year ago

Are you kidding me? 2 upvotes? The OP isn't even using mysql_*().

Michael Berkowski Over a year ago

It's now the correct API, but still not especially helpful. mysqli_fetch_row() returns a numeric indexed array and no matter which is used, the foreach should still return the correct number of columns (though the displayed output from echo "$key\n"; is less useful)

Kumar Gavade Over a year ago

its not the columns but the number of rows that its skipping. The first two always. Please see the new code updated in the question

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