Mysql skips first entry?

[Moopsish]

Hi, First post here

 

I have a problem.. My Mysql database seems to skip the first entry completely..

IE:

you have the database:  

-----------------

ID | entry      |

-----------------

1 | This         |

----------------

2 |   is           |

-----------------

3 |  annoying|

-----------------

 

outputs: is, annoying

 

if it helps here's the part of my code I use to generate/display the data:

$sql = mysqli_query($link, "SELECT * FROM foto");
$row = mysqli_fetch_assoc($sql);
$array = array();




echo "<form method=\"POST\">
<input type=\"hidden\" name=\"id[]\" value=\"".$row['id']."\">";
while ($row = mysqli_fetch_assoc($sql))
{
echo "<input type=\"text\" name=\"foto[]\" value=".htmlspecialchars($row['foto'])."></input>";
}
echo "<input type=\"submit\" name=\"submit3\" value=\"update\"></input></form>";

thanks,

Edited February 3, 2014 by Moopsish

[mac_gyver]

$row = mysqli_fetch_assoc($sql);

 

 

why do you have the above line, at line #2, in the posted code?

[Moopsish]

why do you have the above line, at line #2, in the posted code?

 

Thats so I can echo the Id in line #5. It's the way I learned it.. is there a better way?

Edited February 3, 2014 by Moopsish

[Ch0cu3r]

The while loop wont show the first row because you are calling mysqli_fetch_assoc() on line #2.

This is because each time you fetch a row with  mysqli_fetch_*() it'll return the next row in the  result set.

 

What is the purpose of this code? Is so you can edit multiple records at the same time? If this is the case then you'd setup your form differently. I'd set the record id as the key to foto[]. Example code

$sql = mysqli_query($link, "SELECT * FROM foto");

echo "<form method=\"POST\">";
while ($row = mysqli_fetch_assoc($sql))
{
	echo "<p>#{$row['id']} <input type=\"text\" name=\"foto[{$row['id']}]\" value=".htmlspecialchars($row['foto'])." /></p>";
}
echo "<input type=\"submit\" name=\"submit3\" value=\"update\" />
</form>";

To update all records your code would be

if(isset($_POST['submit3']))
{
	$stmt = mysqli_prepare($link, 'UPDATE foto SET foto=? WHERE id=?');
        mysqli_stmt_bind_param($stmt, 'is', $row_id, $row_value); // bind id and foto value

	foreach($_POST['foto'] as $row_id => $row_value)
	{
		mysqli_stmt_execute($stmt); // update row
	}
}

Edited February 3, 2014 by Ch0cu3r

[Moopsish]

@ch0cu3r thank you so much! the purpuse is indeed to update multiple records. and I've been working on that for awhile now.

thank you for helping me get further into this code

 

-Moopsish