Printing a PHP Array from MySQL table

Question

I am trying to print all the items in a row from a particular table from a mysql database.

Here is the out put I am getting (which is not what I want):

**mysqli_result Object ( [current_field] => 0 [field_count] => 3 [lengths] => [num_rows] => 28 [type] => 0 ) Array ( )**

Here is how I am doing it:

<?php
    $connect = mysqli_connect("localhost", "root", "root");
    mysqli_select_db($connect, "TrackTool");

    $fetch1 = mysqli_query($connect, "SELECT * FROM SalesForceActions");

    // set array
    $array = array();

    // look through query
    while($row = mysql_query($fetch1)){

        // add each row returned into an array
        $array[] = $row;

    }

    print_r($fetch1);
    print_r($array);

    mysqli_close($connect);

?>

What am I doing wrong?

What do you want to obtain? Each row will be stored in an array. — Kneel-Before-ZOD
– Kneel-Before-ZOD, Commented Jul 26, 2013 at 3:05
why are you mixing mysql_ and mysqli_ functions? Your mysqli_ functions are not correct, as it looks like you are using the mysql_ syntax/parameters, but using the mysqli_ prefix. see php.net/manual/en/mysqlinfo.api.choosing.php for examples of each. — Sean
– Sean, Commented Jul 26, 2013 at 3:05
Something that doesn't pan out here. You say that your database name is ActionTable, yet you're trying to connect to TrackTool. Then there's SalesForceActions. Something I'm not getting? — Funk Forty Niner
– Funk Forty Niner, Commented Jul 26, 2013 at 3:19
@Fred yeah i messed up by forgetting to change it in the code.. whoops — Moose
– Moose, Commented Jul 26, 2013 at 3:39

immulatin · Accepted Answer · 2013-07-26 03:07:02Z

5

You need to fetch the result after running the query. So you execute the query then loop which fetching from those results

Like so:

$fetch1 = mysqli_query($connect, "SELECT * FROM SalesForceActions");

$array = array();
while($row = mysqli_fetch_assoc($fetch1)){
    $array[] = $row;
}

You were also mixing mysqli_* functions with mysql_* functions. Be sure to NOT mix those. In fact, just use mysqli_* functions from now on as the mysql_* functions are being deprecated.

answered Jul 26, 2013 at 3:07

immulatin

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Comments

Jason McCreary · Accepted Answer · 2013-07-26 11:13:12Z

3

You need to fetch the result using mysqli_fetch_assoc() (or equivalent):

$array = array();
while($row = mysqli_fetch_assoc($fetch1)) {
  $array[] = $row;
}
print_r($array);

If you indeed want all the rows, take a look mysqli_fetch_all():

$array = mysqli_fetch_all($fetch1, MYSQLI_ASSOC);
print_r($array);

edited Jul 26, 2013 at 11:13

answered Jul 26, 2013 at 2:57

Jason McCreary

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6 Comments

Moose Over a year ago

thats returning the same thing.

Moose Over a year ago

so basically it is returning nothing it looks like this: "Array()"

Jason McCreary Over a year ago

This is not the full code. Replace the relevant code above. If it's still an empty array, then something else is wrong.

Funk Forty Niner Over a year ago

@Moose Another thing I don't get. This answer is the same as you accepted above, minus $fetch1 = mysqli_query($connect, "SELECT * FROM SalesForceActions"); which you already had (posted) and mysql_. Theoretically, this was the first correct answer, or was it the fact that your code didn't work with mysql_? Curious.

Moose Over a year ago

@Fred, its because the other answer pointed out that i messed up by forgetting the "i" in mysqli_*, therefore it was slightly better an answer in my case. Sorry, I can only choose one answer as correct, not two.

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