Pass all variables from mysql/php form

gw32 · July 26, 2011

I need to pass the variable "directory" along with $id. But the more I try the more errors I get, can some one please point me in the correct direction?

echo "<form method=post name=f1 action='somefile.php'>";

echo "<select name='first' onchange=\"reload(this.form)\"><option value=''>Select one</option>";

while($var2 = mysql_fetch_array($quer2)) {
if($var2['DID']==@$id){echo "<option selected value='$var2[DID]'>$var2[directory]</option>"."<BR>";}
else{echo  "<option value='$var2[DID]'>$var2[directory]</option>";}
}
echo "</select>";

This is my first attempt with multiple MySQL/PHP forms.

This selects the "id" from the first table and submits it to a new drop down list using a second table.

The $id passes fine but I also need the variable 'directory' as $path.

Thanks

marcelobm · July 26, 2011

you could do something like:

echo  "<option value='$var2[DID]_$var2[directory]'>$var2[directory]</option>";

and then in the script that receives the value use

$value_arr = explode("_",$value);
and that will give you:

$value_arr[0] = id;
$value_arr[1] = path or directory];

That is the simple solution, the other solution is, when you get the id in the script that process the form, do a query with the id you get to retrieve the directory

gw32 · July 26, 2011

echo  "<option value='$var2[DID]_$var2[directory]'>$var2[directory]</option>";

This would be fine, except that the $var2[directory] breaks the link to the next table lookup. I need it to look like this "firstfile.php?id=1" With your method it appears as "firstfile.php?id=1_pathname" and I get a data error.

Here is the complete code for what I am trying to do.

///////// Getting the data from Mysql table for first list box//////////

$quer2=mysql_query("SELECT DISTINCT directory, DID FROM directory order by directory");  


/////// for second drop down list we will check if directory is selected else we will display all the names///// 

if(isset($id) and strlen($id) > 0){
$quer=mysql_query("SELECT DISTINCT name FROM images where DID=$id order by name"); 
}else{$quer=mysql_query("SELECT DISTINCT name FROM images order by name"); } 

echo "<form method=post name=f1 action='somefile.php'>";

//////////        Starting of first drop down list /////////

echo "<select name='id' onchange=\"reload(this.form)\"><option value=''>Select one</option>";
while($var2 = mysql_fetch_array($quer2)) {
if($var2['DID']==@$id){echo "<option selected value='$var2[DID]'>$var2[directory]</option>"."<BR>";}
else{echo  "<option value='$var2[DID]'>$var2[directory]</option>";}
}
echo "</select>";

//////////        Starting of second drop down list /////////

echo "<select name='name'><option value=''>Select one</option>";
while($var1 = mysql_fetch_array($quer)) { 
echo  "<option value='$var1[name]'>$var1[name]</option>";
}
echo "</select>";

echo "<input type=submit value=Submit>";
echo "</form>";

in somefile.php I have this:

<body>
<?

$id=$_POST['id'];
$image=$_POST['name'];
$path=$_POST['path'];

echo "Value of \$id = $id <br>Value of \$path = $path <br>Value of \$image = $image";
?>

As you can the path variable does not show up.

I could possibly strip the path name from the second table lookup prior to doing the query.

Thanks again.

AyKay47 · July 26, 2011

I don't even see an input with the name of "path" in your code...how would you grab $_POST['path'] if it doesn't exist..? Unless i'm just missing it

gw32 · July 26, 2011

That was my original question, how do I pass the variable '$var2[directory]' from the first drop down list select statement?

Sorry I made it look and sound more difficult than it was supposed to be.

AyKay47 · July 26, 2011

you could do something like:
echo  "<option value='$var2[DID]_$var2[directory]'>$var2[directory]</option>";
and then in the script that receives the value use
$value_arr = explode("_",$value);
and that will give you:

$value_arr[0] = id;
$value_arr[1] = path or directory;
That is the simple solution, the other solution is, when you get the id in the script that process the form, do a query with the id you get to retrieve the directory

why wouldn't this method work for you that he provided?

echo  "<option value='{$var2["DID"]}_{$var2["directory"]}'>{$var2["directory"]}</option>";

fenway · July 26, 2011

I'm not sure why you have multiple values being passed here... what does each one represent?

gw32 · July 26, 2011

Because his solution, although is perfect for another program I am working on, would give me "firstfile.php?id=1_bridal"

When the second sql statement reads $id it is looking for a number (1) the above is not in the table.

the field in the table is DID (int, length 3).

fenway · July 26, 2011

I don't follow -- why do you need both/

AyKay47 · July 26, 2011

Because his solution, although is perfect for another program I am working on, would give me "firstfile.php?id=1_bridal"

When the second sql statement reads $id it is looking for a number (1) the above is not in the table.

the field in the table is DID (int, length 3).

if you are using it for a url then replace the "_" with a "&"

gw32 · July 26, 2011

nope didn't work, I'll try a different method.

Thanks

AyKay47 · July 26, 2011

yeah, i don't get how it get's from an option to a url...you never showed us that...So how can you expect us to help with something that is unclear..try a different method :thumb-up:

gw32 · July 26, 2011

Sir;

My code is this:

///////// Getting the data from Mysql table for first list box//////////

$quer2=mysql_query("SELECT DISTINCT directory, DID FROM directory order by directory");  

//////////        Starting of first drop down list /////////

echo "<select name='id' onchange=\"reload(this.form)\"><option value=''>Select one</option>";
while($var2 = mysql_fetch_array($quer2)) {
if($var2['DID']==@$id){echo "<option selected value='$var2[DID]'>$var2[directory]</option>"."<BR>";}
else{echo  "<option value='$var2[DID]'>$var2[directory]</option>";}
}
echo "</select>";

[/color]

I have this as a passed variable "select name='id'" and also this "$var2['DID']==@$id". Now what I need to pass is this "$var2[directory]".

So how do I do it in it's present context.

The complete code is listed earlier.

Thanks

wildteen88 · July 26, 2011

So you want to retrieve both the $var2['DID'] and $var2['directory'] values from the <select></select> field when the form has been submitted?

gw32 · July 26, 2011

Yes Sir.

wildteen88 · July 26, 2011

The only way to pass both the $var2['DID'] and $var2['directory'] values would be as marcelobm suggested to you earlier

you could do something like:
echo  "<option value='$var2[DID]_$var2[directory]'>$var2[directory]</option>";
and then in the script that receives the value use
$value_arr = explode("_",$value);
and that will give you:

$value_arr[0] = id;
$value_arr[1] = path or directory];
That is the simple solution, the other solution is, when you get the id in the script that process the form, do a query with the id you get to retrieve the directory

To get the id and directory within the script that processes the form you'd use something like this

list($id, $directory) = explode('_', $_POST['id']);

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Pass all variables from mysql/php form

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