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1

I'm trying to make a foreach loop, that runs through the Db table, and echoes it out in a function, but when i try to do this, it takes the rows i have in the object, and echoes the object out as many times as i have rows, and doesn't proceed to the next object in the table. help is appreciated. Here is my code:

function hotels_from_db() {
include 'DbConnection.php';
$sql = "select * from hotels";

$result = $mysqli->query($sql);

$row = $result->fetch_object();

    foreach ($row as $value) {   
    $foundhotel = "<h1>" . $row->hotel_name . "</h1></br>";
    $foundhotel.= $row->hotel_adress . "</br>";
    $foundhotel.= $row->hotel_postal_code . "</br>";
    $foundhotel.= $row->description;
    echo "$foundhotel";
    }  
  }

tried doing this aswell, but this only displays the Last hotel in the table.

function hotels_from_db() {
include 'DbConnection.php';
$sql = "select * from hotels";

$result = $mysqli->query($sql);

$row = $result->fetch_object();
while($row = $result->fetch_object()){           
    $foundhotel = "<h1>" . $row->hotel_name . "</h1></br>";
    $foundhotel.= $row->hotel_adress . "</br>";
    $foundhotel.= $row->hotel_postal_code . "</br>";
    $foundhotel.= $row->description;
    echo "$foundhotel";
    }

}

share|improve this question

4 Answers 4

up vote 0 down vote accepted

fetch_object needs to be called for each row fetch. Try the following:

function hotels_from_db() {
    include 'DbConnection.php';
    $sql = "select * from hotels";

    if ($result = $mysqli->query($sql)) {

        /* fetch object array */
        while ($obj = $result->fetch_object()) {
            $foundhotel = "<h1>" . $obj->hotel_name . "</h1></br>";
            $foundhotel.= $obj->hotel_adress . "</br>";
            $foundhotel.= $obj->hotel_postal_code . "</br>";
            $foundhotel.= $obj->description;
            echo "$foundhotel";        
        }

        /* free result set */
        $result->close();
    }
}
share|improve this answer
    
this return the error - mysqli::query() [mysqli.query]: Empty query, can't explain why though –  Stender Apr 15 '14 at 13:31
    
still returns an error - mysqli_result::fetch_object() expects parameter 1 to be string –  Stender Apr 17 '14 at 15:28
    
Should be fixed. Sorry. Reference: us3.php.net/manual/en/mysqli-result.fetch-object.php –  Amadu Bah Apr 17 '14 at 15:57
    
this did it! thanks –  Stender Apr 19 '14 at 18:08
up vote 2 down vote

->fetch_object retrieves only one row

Try this:

function hotels_from_db() {
include 'DbConnection.php';
$sql = "select * from hotels";
$result = $mysqli->query($sql);

while($row = $result->fetch_object()) {   
    $foundhotel = "<h1>" . $row->hotel_name . "</h1></br>";
    $foundhotel.= $row->hotel_adress . "</br>";
    $foundhotel.= $row->hotel_postal_code . "</br>";
    $foundhotel.= $row->description;
    echo "$foundhotel";
}

}

share|improve this answer
    
hmm, then it only displays the last object in the table, so all the other hotels are not being echoed. –  Stender Apr 15 '14 at 13:24
up vote 0 down vote

It should be:

function hotels_from_db() {
    include 'DbConnection.php';
    $sql = "select * from hotels";

    $result = $mysqli->query($sql);

    $row = $result->fetch_all();

    foreach ($row as $value) {   
        $foundhotel = "<h1>" . $value->hotel_name . "</h1></br>";
        $foundhotel.= $value->hotel_adress . "</br>";
        $foundhotel.= $value->hotel_postal_code . "</br>";
        $foundhotel.= $value->description;
        echo "$foundhotel";
    }  
}
share|improve this answer
    
thanks for the quick comment, but i have tried this, and then it just makes a whole lof of white space, and nothing from the db comes on the screen. –  Stender Apr 15 '14 at 13:20
up vote 0 down vote 
$row = $result->fetch_all(MYSQLI_ASSOC); // MAKE ARRAY ASSOCIATIVE ARRAY

foreach ($row as $value) {   
    $foundhotel = "<h1>" . $value->hotel_name . "</h1></br>";
    $foundhotel.= $value->hotel_adress . "</br>";
    $foundhotel.= $value->hotel_postal_code . "</br>";
    $foundhotel.= $value->description;
    echo "$foundhotel";
}

}

share|improve this answer
    
hmm, this returns a Fatal error: Call to a member function fetch_all() on a non-object, but i see your point, assoc should mean that is checks the next array aswell right? –  Stender Apr 15 '14 at 13:33

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