Hi I'm trying to create a shopping cart for a college assignment, I'm trying to load a table from the contents of a variable but regardless of what I do it won't recognise its contents.
Here the code snippet
$sql = sprintf("SELECT name, description, price FROM %s WHERE id = %d;",$table, $product_id);
$result = mysql_query($sql);
The table variables contents is being missed out so its looking a nameless table, I've searched Google and found a couple of examples but are working for me.
Does anyone have any ideas?
Thanks Scott.
Try this, it should recognize your variable:
$sql = sprintf("SELECT name, description, price FROM " .$table. " WHERE id = %d", product_id);
$result = mysql_query($sql);
mysql_query take 2 parameters, one is your query and other is connection to your database so you need to do the following
$result = mysql_query($sql,your connection variable); `
Try taking off the semicolon after WHERE id = %d
Related
I'm busy with a school project where I need to register users. I created the database and added the tables and can add users. What I just can't get right is to display the next available user id in the table.
I'm using php to retrieve the highest value but when I use echo the variable won't show. There is no error, there is no output at all, just the rest of the page.
Here is the code:
<?php
$db = mysqli_connect('localhost', 'root', '', 'design');
$query = "SELECT MAX(userid) AS userid FROM users" or
die(mysql_error());
$highest_id = mysqli_query($db, $query);
echo $highest_id;
?>
The code successfully connects to the database, the column is called userid, it contains int values and there are other columns as well.
All other code in the script runs perfectly, it's just this part that I can't get to work.
I have spent the last two days reading and searching for answers and I am at my wits end. Any help would be appreciated.
Thank you.
could be your table is User and not Userid
$query = "SELECT MAX(userid) AS userid FROM users"
Anyway for fetching you should use eg:
$result = mysqli_query($db, $query);
$row=mysqli_fetch_array($result,MYSQLI_NUM);
echo $row[0];
The mysqli_query returns a general object that contains the results array. You have to use the mysqli_fetch_row.
<?php
$db = mysqli_connect('localhost', 'root', '', 'design');
$query = "SELECT MAX(userid) AS userid FROM userid" or die(mysql_error());
$highest_id_query = mysqli_query($db, $query);
var_dump($highest_id_query); // so you could check the object attributes
//loop results from query
while($row=mysqli_fetch_row($highest_id_query)){
$highest_id = $row['userid'];
echo $highest_id;
}
?>
You could also use the sql statement: SELECT COUNT(*) FROM userid
Be sure to name your tables correctly! SELECT COUNT(*) FROM users
I'm very new to PHP, trying to figure things out.
I have the following php code:
$read_more = '[read more]';
$sql = "INSERT INTO database (date, headline, article, read_more) VALUES ('$_POST[date]', '$_POST[headline]', '$_POST[article]', '$read_more')";
The code is returning "http://www.example.com/index.php?id=0". Note that the "id" parameter is returning "0". My goal is to make it return the latest ID from the database, which is set to auto increment.
I've tried many things but nothing worked for me so far. Thanks!
EDIT: After hours of trial and error, this is how I was able to solve this problem:
//After connecting to the database
$sql = "INSERT INTO table (date, headline, article, read_more) VALUES ('$_POST[date]', '$_POST[headline]', '$_POST[article]', '$read_more')";
$result = mysqli_query($conn, $sql);
$id = mysqli_insert_id($conn);
Now the latest id is saved into a variable that I can use.
Try something like this instead. Check the documentation here.
// insert a datarow, primary key is auto_increment
// value is a unique key
$query = "INSERT INTO test (value) VALUES ('test')";
mysql_query( $query );
echo 'LAST_INSERT_ID: ',
mysql_query( "SELECT LAST_INSERT_ID()" ),
'<br>mysql_insert_id: ',
mysql_insert_id();
I am trying to use a mssql Query to retrieve an id from a table row.
This is the code I am trying to use:
$username = $_COOKIE['ID_my_site'];
$sql ="SELECT id FROM users WHERE username = ".$username."";
$query = mssql_query($sql, $conn);
$array = mssql_fetch_assoc($query);
$acc_id=stripslashes($array['id']);
var_dump ($sql);
echo '<script>
alert("'.$acc_id.'");
</script>';
When I use this though, the sql is correct, from what I see using var_dump. But the alert from JavaScript is blank.
How can I get the alert to display the data from the id column?
The id data should be 5.
Thank you for any help, all help is appreciated.
If you +1 my question I will +1 your answer.
I will +1 an answer if I choose it as best answer, regardless if you +1 my question or not!
Missing quotes in SQL query, try following
$sql ="SELECT id FROM users WHERE username = '".$username."'";
^ quotes ^
OK So I'm trying to access a table called emg_quote I have the Quote ID so Im trying to get the Column Subject from the same row as this ID but for some reason All I'm getting is the first row in the entire table? Can any one figure out what I'm doing wrong? Here is my coding:
$row['quote_id'] = quoteTitle($row['quote_id']);
function quoteTitle($quoteid){
global $db;
$sql = "SELECT subject FROM emg_quote WHERE ".$quoteid."";
$res = $db->query($sql);
$row = $db->fetch_row();
$output = $row['subject'];
return $output;
}
Are you using a custom object to wrap the native API's?
Either way it doesn't look right to me. You don't seem to be using the result of the query.
i.e.
$result = $mysqli->query($query);
$row = $result->fetch_row();
You have few bad practices in your code.
A. You lie on $quoteid to give you the correct where syntax. ie: ID=123
This is an highly unsafe method, because the user can change the it to Some-Important-Details='bla'
To extract more details from this table or others.
B. You should ALWAYS escape characters when receiving data from user, otherwise you easily subjected to SQL-Injections. And believe me you don't want it.
you have to use the checking after where.
use you column name before your $quoteid variable
$row['quote_id'] = quoteTitle($row['quote_id']);
function quoteTitle($quoteid){
global $db;
$sql = "SELECT subject FROM emg_quote WHERE quoteid=".$quoteid." LIMIT 1 ";
$res = $db->query($sql);
$row = $db->fetch_row();
$output = $row['subject'];
return $output;
}
Remember : USE limit 1 when you search with primary key and you know that only 1 record will be searched. it reduce your processing time.
You might be missing the where column.
$sql = "SELECT subject FROM emg_quote WHERE quote_id=".$quoteid."";
^^^^^^^^
We also do not see weather something with your Db class is wrong.
You should in any case not directly put request variables into a database query.
$sql = "SELECT subject FROM emg_quote WHERE ID='".$quoteid."'";
You had not wrote your db fieldname in where condition
PLease note I am a beginner.
My situation is thus:
I am trying to run multiple queries, off the back of a dynamic form. So the data is going to end up in two different tables.
I am currently successfully storing in to my item_bank, which has an auto_increment itemId.
I then want to grab the ItemId just created on that last query and insert it into my next query of which I am also inserting an array. (I hope you can follow this)
first off, is it even possible for me to run multiple queries like this on a single page?
Below is my attempt at the queries. Currently the first query works, however I cannot get the ItemId generated from that query.
$answers is an array.
// store item structure info into item_bank_tb
$query = "INSERT INTO item_bank_tb (item_type, user_id, unit_id, question_text, item_desc, item_name)
VALUES('$type','$creator','$unit','$text','$desc','$name')";
mysql_query($query) or die(mysql_error());
$itemid = mysql_insert_id();
//now store different answers
$query = "INSERT INTO answers_tb (item_id, text_value)VALUES('$itemid',' . implode(',', $answers) . ')";
mysql_query($query) or die(mysql_error());
this is the error i get now: "Column count doesn't match value count at row 1"
To get the ID generated by an auto-increment field in PHP/MySQL just use the mysql_insert_id() function.
Edit:
// store item structure info into item_bank_tb
$query = "INSERT INTO item_bank_tb (item_type, user_id, unit_id, question_text, item_desc, item_name)
VALUES('$type','$creator','$unit','$text','$desc','$name')";
mysql_query($query) or die(mysql_error());
$itemid = mysql_insert_id();
//now store different answers
$answers = mysql_real_escape_string(implode(',', $answers));
$query = "INSERT INTO answers_tb (item_id, text_value) VALUES('$itemid','$answers')";
mysql_query($query) or die(mysql_error());
Have a look at mysql_insert_id():
// store item structure info into item_bank_tb
mysql_query($query);
$itemid = mysql_insert_id();
// now store different answers