Hello everyone, this is my first post here so I will try to make it as clear as possible.
Firstly i have a database with a single table named "albums" with the following fields: -
albumid int(11)
-
genreid varchar(30)
-
albumtitle varchar(30)
-
albumauthor varchar(30
-
numberofsongs int(11
-
price int(11)
Then i have two php pages
Search.php : -
<html>
-
-
<center>
-
-
</br></br></br></br>
-
-
<img src="guitarlogo.jpg" alt="Music Database"/>
-
-
<form method="GET" action="Result.php">
-
-
<h2><b>Search Music Database:</b></h2>
-
-
<input type="text" name="mts" />
-
<input type="submit" name="submit" value="Search" />
-
</form>
-
-
<?php
-
$connection = mysql_connect("localhost","root","");
-
$fields = mysql_list_fields("ea09039", "albums", $connection);
-
$columns = mysql_num_fields($fields);
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echo "<form action=Result.php method=POST>";
-
-
echo "<select name=Field>";
-
for ($i = 0; $i < $columns; $i++) {
-
echo "<option value=$i>";
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echo mysql_field_name($fields, $i);
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}
-
echo "</select></form>";
-
?>
-
-
</center>
-
-
</html>
-
and Result.php : -
<?php
-
-
include("db.php");
-
-
$mts=$_GET["mts"];
-
$Field=$_GET["Field"];
-
-
-
$query="select * from albums where '$Field' like '%$mts%'";
-
-
$res2=mysql_query($query);
-
-
while($row=mysql_fetch_row($res2))
-
{
-
echo $row[2];
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echo " has author ";
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echo $row[3];
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echo " and costs ";
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echo $row[5];
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echo " Euro.";
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echo " And is from the genre ";
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echo $row[1];
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echo "<br/>";
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}
-
-
?>
-
This is as far as i have gotten with this, but i keep getting errors.
Please if someone can help I would be very thankful.
12  7084  JKing 1,206
Recognized Expert Top Contributor
Okay, so what are your errors? What is it doing? What is it supposed to be doing?
Right now it's just giving me a blank page.
I wrote the word "blue" in the textbox and selected "albumtitle" from the dropdown list(which is dynamically popullated from a table in a MySQL database).
This is what i get at the address bar after i do a search: - http://localhost/ea09039/HOMEWORK2/Result.php?mts=blue&submit=Search
As for what is it supposed to be doing,
I need it to take the input from the textbox, and search within the column from the table that is selected from the dropdown list which are in Search.php and show the results in Result.php.
I hope i was clear :)
JKing 1,206
Recognized Expert Top Contributor
It looks like you have two forms on your search.php page. The first form only has a text input for mts and a submit button. These are sent to result.php via GET.
Your second form has a select named field and this form is sent to result.php via POST however there is no submit button for this form.
On result.php you try to retrieve the field variable via $_GET but you have never sent it with GET or at all because it is part of another form that doesn't get submitted and if it were sent it would be in the $_POST.
Now since your $field variable is never populated your query fails and so does your while loop and you end up with no output.
The solution here is to make the select part of your first form.
Also use mysql_real_escape_string() on any variables you pass to the database.
Okay thank you for the reply.
I restructured my code and I believe there is one more problem left.
Here is the code for Search.php -
<html>
-
-
<center>
-
-
</br></br></br></br>
-
-
<img src="guitarlogo.jpg" alt="Music Database"/>
-
-
<h2><b>Search Music Database:</b></h2>
-
<?php
-
$connection = mysql_connect("localhost","root","");
-
$fields = mysql_list_fields("ea09039", "albums", $connection);
-
$columns = mysql_num_fields($fields);
-
-
echo "<form action=Result.php method=GET>";
-
echo "<input type=text name=mts/>";
-
echo "<select name=Field>";
-
-
for ($i = 0; $i < $columns; $i++) {
-
echo "<option value=$i>";
-
echo mysql_field_name($fields, $i);
-
}
-
echo "</select>";
-
echo "<input type=submit name=submit value=Search />";
-
?>
-
-
-
</center>
-
-
</html>
-
The problem is that the Field variable returns the number of the position on the table. For example If i select albumtitle from the dropdown the Field variable return the position of the column albumtitle which is 2. I need it to return the name of that column not it's position.
Here is the result of the address barr - http://localhost/ea09039/HOMEWORK2/Result.php?mts/=blue&Field=2&submit=Search
Here is the Results.php file again: -
<?php
-
-
include("db.php");
-
-
$mts=$_GET["mts"];
-
$Field=$_GET["Field"];
-
-
- $query="select * from albums where '$Field' like '%$mts%'";
-
-
$res2=mysql_query($query);
-
-
while($row=mysql_fetch_row($res2))
-
{
-
echo $row[2];
-
echo " ka autor ";
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echo $row[3];
-
echo " dhe kushton ";
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echo $row[5];
-
echo " Euro.";
-
echo " Dhe eshte nga zhanri ";
-
echo $row[1];
-
echo "<br/>";
-
}
-
-
?>
-
Any ideas?
Not what I'm looking for, but it might come in handy in the future. Thanks anyway
JKing 1,206
Recognized Expert Top Contributor
The problem is that the Field variable returns the number of the position on the table. For example If i select albumtitle from the dropdown the Field variable return the position of the column albumtitle which is 2. I need it to return the name of that column not it's position.
Well of course it is. You are setting all your option values equal to $i in your for loop and not the field name.
Okay thank you i understand my mistake, but could you please show me how should the code look, i can not seem to figure it out.
P.S im new to php so please have patience with me :)
- $_id = $_POST['Field'];
-
-
while($row=mysql_fetch_row($res2)) {
-
if ( $row['id'] == $_id ) {
-
$_name = $row['name'];
-
}
-
}
Or something like that. Basicly compare the id from the $_POST variable or instead of that... - $connection = mysql_connect("localhost","root","");
-
$fields = mysql_list_fields("ea09039", "albums", $connection);
-
$columns = mysql_num_fields($fields);
-
-
echo "<form action=Result.php method=GET>";
-
echo "<input type=text name=mts/>";
-
echo "<select name=Field>";
-
-
for ($i = 0; $i < $columns; $i++) {
- echo "<option value='". mysql_field_name($fields, $i). "'>";
-
echo mysql_field_name($fields, $i);
-
}
-
echo "</select>";
-
echo "<input type=submit name=submit value=Search />";
Thank you Samishii23 the second part worked like a charm :D
Also thanks to everyone that contributed
Let me make a note about doing that. When you pass data like this there is a higher chance of someone finding out what your data is because they know something about your Database, though it is a bit far fetched, but the concept and possibility is there. Your better off using the #'s rather then the Field name itself.
You could use a field name array and do something like this: - $fields = array ('id','name','date');
-
$Field = $fields[$_POST['Field']];
Thank you for the information, i will try it out.
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