Using for loop instead of foreach throws an error

Question

I am using a simple for loop to print field names of a table and it didn't work but the same code when I wrote using foreach loop it worked perfectly fine. I am a bit confused as to what might have caused it. Please help me find where I went wrong with the coding.

Code using foreach-loop:

$data=mysqli_query($con,"select * from `address` order by $id $sort");
echo "<tr>";
    $field=mysqli_fetch_fields($data);
    foreach ($field as $val) {
        echo "<th><a href='sort_table_data_asc_desc.php?fn=$val->name &ord=$sort'>$val->name</a></th>";
    }
echo "</tr>";

Output: as requested

Code using for-loop:

$data=mysqli_query($con,"select * from `address` order by $id $sort");
$col=mysqli_num_fields($data);
echo "<tr>";
    $field=mysqli_fetch_fields($data);
    for ($i=0;$i<$col;$i++) {
        echo "<th><a href='sort_table_data_asc_desc.php?fn=$field->name &ord=$sort'>$field->name</a></th>";
    }
echo "</tr>";

Output: as requested

Can you edit to show the expected output and what you actually got? — Nick
– Nick, Commented Dec 19, 2017 at 19:36

Nigel Ren · Accepted Answer · 2017-12-19 19:44:45Z

2

mysqli_fetch_fields returns an array of the fields, so you need to reference the array element when accessing the values

echo "<th><a href='sort_table_data_asc_desc.php?fn={$field[$i]->name}&ord=$sort'>{$field[$i]->name}</a></th>";

As you can see, I'm using $field[$i] at each point you access the fields coming back from mysqli_fetch_fields

edited Dec 19, 2017 at 19:44

answered Dec 19, 2017 at 19:38

Nigel Ren

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6 Comments

Don'tDownvoteMe Over a year ago

I did what you said but it throws an error, "Object of class stdClass could not be converted to string"

Nigel Ren Over a year ago

I've updated it to include {} round the field, it makes the field itself clearer, can you try that

Nigel Ren Over a year ago

I've just tried it on a table I have and it works OK, have you changed both references to $field?

Don'tDownvoteMe Over a year ago

Yes @Nigel, I changed both the references in the URL as you said but it didn't work on my end.

Nigel Ren Over a year ago

If the error is giving a particular line of code that's showing the error, can you make sure that is in the code on your original question.

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