Insert statment only insert one field into new table - php

I'm having a issue with an insert statement. It should transfer a row of data from one table to another upon clicking a link from a table from that specified row.
At the minute the only field being inserted into the other table is the reference.
In other words I want to retrieve a person's name and details from a row in one table which corresponds to a unique reference number and move it to the next table in the process.
Table Page.php
<td align="center">Refer for Triage</td>
<td align="center">Refer for IA</td>
Script.php - action page
require_once('auth.php');
$host="localhost"; // Host name
$username="root"; // Mysql username
$password="sysadmin"; // Mysql password
$db_name="Elite"; // Database name
$tbl_name="Instruction"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$Reference=$_REQUEST['Reference'];
$Forename=$_REQUEST['Forename'];
$surname=$_REQUEST['surname'];
$DOB=$_REQUEST['DOB'];
$Mobile=$_REQUEST['Mobile'];
$Home=$_REQUEST['Home'];
$Address=$_REQUEST['Address'];
$Postcode=$_REQUEST['Postcode1'];
$Email=$_REQUEST['Email'];
$Accident=$_REQUEST['Accident'];
$Details=$_REQUEST['Details'];
//semi colon removed
$sql="INSERT INTO
Triage
(
Reference,
Forename,
surname,
`D.O.B`,
`Mobile Number`,
`Home Number`,
Address,
Postcode1,
Email,
Accident,
Details
)
VALUES
(
'".$Reference."',
'".$Forename."',
'".$surname."',
'".$DOB."',
'".$Mobile."',
'".$Home."',
'".$Address."',
'".$Postcode1."',
'".$Email."',
'".$Accident."',
'".$Details."'
)";
$result=mysql_query($sql);
// echo "Successful";
// echo "<BR>";
// echo "<a href='list_records.php'>View result</a>";
// mysql_error()
echo $sql
?>

Related

INSERT into two different tables with a related key

I'm trying to insert into several related database tables some data after clicking a submit button, my tables are:
USERS: ID (primary key), User, Name, Password
LEVELS: ID, User_ID (foreign key), Level1, Level2, Level3, Level4
where User_ID on the table levels is the same ID as the primary key for users.
I want to make this insert with php, my code is as follows:
$host="xxxxxx"; // Host name
$username="xxxxxx"; // Mysql username
$password="xxxxxx"; // Mysql password
$db_name="xxxxxx"; // Database name
$tbl_name="USERS"; // Table name
// Connect to server and select databse.
$dbh= mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// sent from form
$name=$_POST["name"];
$user=$_POST["user"];
$password=$_POST["password"];
$L4=$_POST["L4"];
$L3=$_POST["L3"];
$L2=$_POST["L2"];
$L1=$_POST["L1"];
$sql="INSERT INTO $tbl_name (Name, User, Password) VALUES('$name','$user','$password');";
$userid = mysql_insert_id();
$tab1= mysql_query($sql, $dbh) or die ("problem query 1");
$sql2 = "INSERT INTO LEVELS (User_ID, Level1, Level2, Level3, Level4) VALUES('$userid','$L1','$L2','$L3','$L4');";
$tab2= mysql_query($sql2, $dbh) or die ("problem query 2");
either, I don't get how to relate the tables, or something here is wrong, cause only the first sql statement is being executed, and the second one prints the die 'problem query 2'.
Can anybody please help me?
Thanks!
$userid = mysql_insert_id();
should be called after the insert query is executed and in your case you are calling it before the first query being executed.
So it should be as
$sql="INSERT INTO $tbl_name (Name, User, Password) VALUES('$name','$user','$password');";
$tab1= mysql_query($sql, $dbh) or die ("problem query 1");
$userid = mysql_insert_id();

Insert statement not working. not transferring into other table

Pretty new to PHP and MySQL.
I have created an insert statement in my php script, to transfer a row of data from one table to the next for certain fields. Only thing is, it doesn't seem to be working?
Can anybody see where the issue is?
<?php
require_once('auth.php');
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="Instruction"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details);
VALUES (Reference,Forename,surname,DOB,Mobile,Home,Address,Postcode1,Email,Accident,Details)";
$result=mysql_query($sql);
//
while($rows=mysql_fetch_array($result)){
echo 'update test';
}
//
// end of while loop
echo "Successful";
echo "<BR>";
echo "<a href='list_records.php'>View result</a>";
?>
Update
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$Reference=$_REQUEST['Reference'];
$Forename=$_REQUEST['Forename'];
$surname=$_REQUEST['surname'];
$DOB=$_REQUEST['DOB'];
$Mobile=$_REQUEST['Mobile'];
$Home=$_REQUEST['Home'];
$Address=$_REQUEST['Address'];
$Postcode=$_REQUEST['Postcode1'];
$Email=$_REQUEST['Email'];
$Accident=$_REQUEST['Accident'];
$Details=$_REQUEST['Details'];
//semi colon removed
$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details)
VALUES('.$Reference.','.$Forename.','.$surname.','.$DOB.','.$Mobile.','.$Home.','.$Address.','.$Postcode1.','.$Email.','.$Accident.','.$Details.')";
$result=mysql_query($sql);
echo "Successful";
echo "<BR>";
echo "<a href='list_records.php'>View result</a>";
?>
first you should fix the assignments:
$Reference=$_REQUEST['Reference'];
$Reference=$_REQUEST['Forename'];
...
should be something like:
$Reference=$_REQUEST['Reference'];
$Forename=$_REQUEST['Forename'];
$surname=$_REQUEST['surname'];
Then update the query in:
$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details)
VALUES (".$Reference.",".$Forename.","...
and so on with the rest of the values.
Also
while($rows=mysql_fetch_array($result)){
won't work since result will only contain true on success.
Maybe there are more mistakes I'm not sure. But you should also check this to learn how to avoid injection:
What's the best method for sanitizing user input with PHP?
If you want to transfer data from one table to another, you should select this table somewhere. You have not anywhere in your code, you just specified columns, how is your script supposed to know where do they come from?
INSERT INTO table1 (col1, col2, col3) SELECT correspondingColumn1, correspondingColumn2, correspondingColumn3 FROM table2
P.S.: You do not use $Reference, but still, you are overwritting it
try this one
1) you mention all var name as $Reference its changed
2) query not correct plz study how wrote query..
3) REFER:http://www.w3schools.com/php/php_mysql_intro.asp
<?php
require_once('auth.php');
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="Instruction"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$Reference=$_REQUEST['Reference'];
$Forename=$_REQUEST['Forename'];
$surname=$_REQUEST['surname'];
$DOB=$_REQUEST['DOB'];
$Mobile=$_REQUEST['Mobile'];
$Home=$_REQUEST['Home'];
$Address=$_REQUEST['Address'];
$Postcode=$_REQUEST['Postcode1'];
$Email=$_REQUEST['Email'];
$Accident=$_REQUEST['Accident'];
$Details=$_REQUEST['Details'];
//semi colon removed
$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details)
VALUES ('$Reference','$Forename','$surname','$DOB','$Mobile','$Home','$Address','$Postcode1','$Email','$Accident','$Details')";
$result=mysql_query($sql);
//
while($rows=mysql_fetch_array($result)){
echo 'update test';
}
//
// end of while loop
echo "Successful";
echo "<BR>";
echo "<a href='list_records.php'>View result</a>";
?>

Insert statement not working. transfer from one table to another

Pretty new to PHP and MySQL.
I have created an insert statement in my php script, to transfer a row of data from one table to the next for certain fields. Only thing is, it doesn't seem to be working?
Can anybody see where the issue is?
<?php
require_once('auth.php');
$host=""; // Host name
$username=""; // Mysql username
$password=""; // Mysql password
$db_name=""; // Database name
$tbl_name="Instruction"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$Reference=$_REQUEST['Reference'];
$Forename=$_REQUEST['Forename'];
$surname=$_REQUEST['surname'];
$DOB=$_REQUEST['DOB'];
$Mobile=$_REQUEST['Mobile'];
$Home=$_REQUEST['Home'];
$Address=$_REQUEST['Address'];
$Postcode=$_REQUEST['Postcode1'];
$Email=$_REQUEST['Email'];
$Accident=$_REQUEST['Accident'];
$Details=$_REQUEST['Details'];
//semi colon removed
$sql="INSERT INTO Triage (Reference,Forename,surname,D.O.B,Mobile Number,Home Number,Address,Postcode1,Email,Accident,Details)
VALUES('.$Reference.','.$Forename.','.$surname.','.$DOB.','.$Mobile.','.$Home.','.$Address.','.$Postcode1.','.$Email.','.$Accident.','.$Details.')";
$result=mysql_query($sql);
echo "Successful";
echo "<BR>";
echo "<a href='list_records.php'>View result</a>";
?>
You have column names like Mobile Numbe etc, you need to use `` for them also the concatenation does not look correct you should have something as
sql="INSERT INTO
Triage
(
Reference,
Forename,
surname,
`D.O.B`,
`Mobile Number`,
`Home Number`,
Address,
Postcode1,
Email,
Accident,
Details
)
VALUES
(
'".$Reference."',
'".$Forename."',
'".$surname."',
'".$DOB."',
'".$Mobile."',
'".$Home."',
'".$Address."',
'".$Postcode1."',
'".$Email."',
'".$Accident."',
'".$Details."'
)";
In addition you should use mysql_real_escape_string() for all the request data something as
$Reference=mysql_real_escape_string($_REQUEST['Reference']);
and so on for others

Form submitting creating an 500 error

I am trying to make this form work. But when I click submit it goes to at 500 internal server error.
Am I doing something wrong here or am I way off?
I've tried everything that was obvious changed post action even tried changing the submit button action.
This is my script for the form action.
<?php
$host="localhost"; // Host name
$username="signup"; // Mysql username
$password="pass"; // Mysql password
$db_name="signup"; // Database name
$tbl_name="signup_mysql"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// Get values from form
$venue_name=$_POST['venue_name'];
$production_company_name=$_POST['production_company_name'];
$venue_contact_name=$_POST['venue_contact_name'];
$location=$_POST['location'];
$office_location=$_POST['office_location'];
$capacity=$_POST['capacity'];
$age=$_POST['age'];
$description=$_POST['description'];
$facebook=$_POST['facebook'];
$twitter=$_POST['twitter'];
$soundcloud=$_POST['soundcloud'];
$website=$_POST['website'];
// Insert data into mysql
$sql="INSERT INTO $tbl_name(venue_name, production_company_name, venue_contact_name, email, location, office_location, capacity, age, description, facebook, twitter, soundcloud, website)VALUES('$venue_name', '$production_company_name', '$venue_contact_name', '$email', '$location', '$office_location', '$capacity', '$age', '$description', '$facebook', '$twitter', '$soundcloud', '$website')";
$result=mysql_query($sql);
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='index.html'>Back to main page</a>";
}
else {
echo "ERROR";
}
?>
<?php
// close connection
mysql_close();
?>

Hostgator SQL database not updating with PHP form

I have a PHP form that should insert data into my SQL database on hostgator. However it is not adding any data but the id field keeps incrementing. I do not receive any error message when submitting the form and when i go to the database the other fields are just empty thus not displaying any data.
I am pulling my hair and cant figure out what the problem is. Can someone please help me
Thanks
<?php
$host="localhost"; // Host name
$username="xxxxxx"; // Mysql username
$password="xxxxxx"; // Mysql password
$db_name="rob1124_inventory"; // Database name
$tbl_name="data"; // Table name
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
// Get values from form
$qty=$_POST['qty'];
$product=$_POST['product'];
$price=$_POST['price'];
$totalprice=$_POST['totalprice'];
$seller=$_POST['seller'];
$city=$_POST['city'];
// Insert data into mysql
$sql="INSERT INTO $tbl_name(qty, product, price, totalprice, seller,city)
VALUES('$qty', '$product', '$price', '$totalprice', '$seller', '$city')";
$result=mysql_query($sql);
// if successfully insert data into database, displays message "Successful".
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='insert.php'>Back to main page</a>";
}
else {
echo "ERROR";
}
?>
<?php
// close connection
mysql_close();
?>
Change to utf-8 from all varchar fields of your table and
try to get mysql_error().
// Connect to server and select database.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
mysql_query("set names 'utf8'");
//You codes....
// Insert data into mysql
$sql="INSERT INTO $tbl_name(qty, product, price, totalprice, seller,city)
VALUES('$qty', '$product', '$price', '$totalprice', '$seller', '$city')";
$result=mysql_query($sql) or die(mysql_error());
//Your codes...
Since the id is incrementing atleast the form and the DB connect, it tries to enter data.
One usually occurring error is that the data types in the databases columns don't match with the type of data recieved. Like trying to insert chars into ints etc. Or the length of the data is to large for the assigned size in the database. Check to see that the types are correct and try again.
But still, those that are correct should be inserted. Hard to tell without knowing more about the database design.

Categories