I have searched high and low and cannot find anything to fix this problem. apparently my first insert statement runs fine and I can see the changes in the database. The second one fails to update the database even though the script completes. Any help would be greatly appreciated. Thanks in advance.
"INSERT INTO propertyInfo (
districtNumber,
primaryProperty,
platBookPage,
giftedDate,
purchaseDate,
retiredDate,
yearBuilt,
Affiliation,
purchasePrice,
primaryZone,
primaryLandUse,
propertyType,
squareFootage,
soldDate,
soldPrice,
lotSizeSqFt,
iudb,
zoningLandUse,
enterpriseZone,
zoningCode,
florida,
georgia,
Description,
Land,
Development,
acres,
tax,
Name,
granter
)
VALUES (
'$district_number',
'$primary_property',
'$plat_book_page',
'$gifted_date',
'$purchase_date',
'$retired_date',
'$year_built',
'$affiliation',
'$purchase_price',
'$primary_zone',
'$primary_land_use',
'$property_type',
'$square_footage',
'$sold_date',
'$sold_price',
'$lot_size_sqft',
'$iudb',
'$zoning_land_use',
'$enterprise_zone',
'$zoning_code',
'$florida',
'$georgia',
'$description',
'$land',
'$development',
'$acres',
'$tax',
'$name',
'$granter'
)";
$pinksql = "INSERT INTO address (`address`, `address2`, `city`, `state`, `zipCode`, `subDivision`) VALUES ('$address', '$address2', '$city', '$state', '$zip_code', '$sub_division')";
Related
$logquery="INSERT INTO new_data_entry_log (directory,ip,description,state,status,instituteid,email,category,bandwidth,project,institutehead,contactnumber,landlinenumber,latitude,longitude,clientip,added by,added on,added at) VALUES ('$directory','$ipaddress','$description','$state','$status','$instituteid','$contactemail','$bandwidth','$institutehead','$contactNumber','$contactlandlinenumber','$latitude','$longitude','$clientip','$name','$currentdate','$currenttime')";
I can see there is a mismatch of columns count and related values count. Probably it might be the issue.
Did you try with removing the single quates of the variables?
I have formated your query.
Formated Query
$logquery="INSERT INTO `new_data_entry_log`(
`directory`,
`ip`,
`description`,
`state`,
`status`,
`instituteid`,
`email`,
`category`,
`bandwidth`,
`project`,
`institutehead`,
`contactnumber`,
`landlinenumber`,
`latitude`,
`longitude`,
`clientip`,
`added by`,
`added on`,
`added at`
) VALUES(
'$directory',
'$ipaddress',
'$description',
'$state',
'$status',
'$instituteid',
'$contactemail',
'$bandwidth',
'$institutehead',
'$contactNumber',
'$contactlandlinenumber',
'$latitude',
'$longitude',
'$clientip',
'$name',
'$currentdate',
'$currenttime'
)";
In your code
new_data_entry_log(19 property)
But
VALUES(17 values)
This is the main reason. Another reason can be data type issue.
Because you set all values as String/Varchar
Note: best practice(underscore) for database fields
`added by` --> 'added_by'
`added on` --> 'added_on'
`added at` --> 'added_at'
I am using a simple php script to insert data into database but it's failing. The query just doesn't become successful without showing a single error which is why I am unable to figure out the problem. Some expert here help me please.
echo $name." ".$email." ".$pass." ".$phone." ".$area." ".$specialization." ".$city." ".$latitude." ".$longitude;
The result of echo is normal - without any null elements.
$query = mysqli_query($conn, "INSERT INTO users (name, email, pass, phone, area, specialization, hospital, city, latitude, longitude)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ");
if ($query) {
echo "Status: Registeration Successful!";
// creating directory for user and storing dummy profile picture
//mkdir('../profiles/'.$email_trim, 0777);
//$result_copy = copy("img/dp.jpg.jpg", "../profiles/".$email_trim."/dp.jpg.jpg");
} else {
echo "Status: Err";
}
This "Status: Err" is always printed. I don't know why.
P.S I have double checked the database the field labels are fine.
UPDATE 1:
I added the
die(mysqli_error($conn));
statement and it says "DUPLICATE ENTRY '0' FOR KEY PRIMARY'.
PROBLEM AND SOLUTION:
The issue was that I had an 'id' field which was primary key of the table but it was not set to AUTO_INCREMENT. So, whenever I tried to insert a new record, I was actually inserting entries with duplicate PKs which was the issue. I change it to AUTO_INCREMENT and it solved the problem.
It seems you try to insert a new element with a PK = 0, but there is already a record with this key !
What is the primary key of your table ? Do you use an "id" field which is not shown in your insert statement ? Is this field AUTO_INCREMENT ?
It would be helpful to see the structure of your 'users' table.
Wild guess: looks like you may have defined an "id" column (or with whatever other name) which is primary key with default value "0", but it's not auto increment. That way you can insert 1 row and it will get "0" as "id" column's value, but you cannot insert another row because it will also try to use default value "0", which cannot happen as primary key has to be unique.
If that is the case, then please alter users table and make sure that the primary key column is also 'auto increment'.
Please check the proper error by adding below mentioned code inside else:
echo mysqli_errno($conn) . '----' . mysqli_error($conn);
<?php
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
$query = mysqli_query($conn, "INSERT INTO `users` (`name`, `email`, `pass`, `phone`, `area`, `specialization`, `hospital`, `city`, `latitude`, `longitude`)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ") or die(mysqli_error());
$query = mysqli_query($conn, "INSERT INTO `users` (`name`, `email`, `pass`, `phone`, `area`, `specialization`, `hospital`, `city`, `latitude`, `longitude`)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude') ");
use ` Tick maybe because there's some reserved word in your fields.
I think you doubled your close parenthesis and do not put $conn inside the query..
$query = "INSERT INTO users (name, email, pass, phone, area, specialization, hospital, city, latitude, longitude)
VALUES ('$name', '$email', '$pass', '$phone', '$area', '$specialization', '$hospital', '$city', '$latitude', '$longitude')";
mysqli_query($query, $conn);
Something like this. I hope this helps
I will admit I am a newbie when it comes to PDO, but I have to change over a form that is in mysql.. I am getting connection, but nothing inserted.. I am truly stuck and feel like an idiot because I know it is something simple I am missing
I have tried having the arrays above and after the insert.. Neither work
Any help would be appreciated
Here is my code:
$conn = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);
$STH = $conn->prepare("INSERT INTO PinTrade (ID, PIN, Year, Make, Model, Mileage, FirstName, LastName, Phone, Email, Date)
VALUES ('', '$pin', '$year', '$make', '$model', '$mileage', '$first', '$last', '$phone', '$email', '1234' )");
$STH->bindParam(':PIN', $_POST['pin']);
$STH->bindParam(':Year', $_POST['year']);
$STH->bindParam(':Make', $_POST['make']);
$STH->bindParam(':Model', $_POST['model']);
$STH->bindParam(':Mileage', $_POST['mileage']);
$STH->bindParam(':FirstName', $_POST['first']);
$STH->bindParam(':LastName', $_POST['last']);
$STH->bindParam(':Phone', $_POST['phone']);
$STH->bindParam(':Email', $_POST['email']);
$STH->execute();
Get rid of the dollar signs and quotes in your query values:
$STH = $conn->prepare("INSERT INTO PinTrade (ID, PIN, Year, Make,
Model, Mileage, FirstName, LastName, Phone, Email, Date)
VALUES (null, :PIN, :Year, :Make, //and so on....
Also note, assuming ID is an auto incrementing field, just insert null
VALUES (null, :PIN,
Finally, if you're pulling from the post array, I'd use bindValue over bindParam
I am working on building an employee database but seem to have run into an interesting issue.
When I run my query to add a new user/employee, I get the error:
Column count doesn't match value count at row 1
From what I have researched, this seems to be an error with inserting more/less values than what is declared in the first part of an insert statement example:
INSERT INTO table (col1, col2) VALUES (val1, val2, val3)
The thing is though, I have looked over my query and the columns and values match perfectly (count wise). I have even looked for things in my query such as missing quotes, commas, etc.
Here is my code (query):
$db->query("INSERT INTO userdata (
Username,
Email,
Phone,
Password,
FirstName,
LastName,
Address,
City,
State,
Zip,
JobTitle,
UserGroup,
JobStatus,
Points,
Status,
BirthDate,
HireDate,
HourlyRate,
DateAdded,
SSN
) VALUES (
'$Username',
'$Email',
'$Phone',
'$Password',
'$FirstName',
'$LastName',
'$Address',
'$City',
'$State',
'$Zip',
'$JobTitle',
'$Group',
'$JobStatus',
0,
'$Status',
'$BirthDate',
'$HireDate',
'$HourlyRate'
'$TodaysDate',
'$SSN'
)") or die(mysqli_error($db));
Some things to note:
This not all of the columns in the table have data inserted here (I think its possible to do this and things such as auto incrementing ID's will fill themselves in and others will be left blank)
From the variable dumps I have done, all of these variables are valid.
I am really confused about this and any help would be appreciated.
Check the following portion again:
INSERT INTO userdata(...,
JobStatus,
UserGroup,
Points,
UserGroup,
Status,
.....
,)VALUES(...,
$JobStatus',
0,
'$Group',
'$Status',
......
)
In values(, ?, ?, ?), after jobstatus, there should be UserGroup and then Points. And UserGroup appeared twice.
I am really having a tough time understanding what syntax to use. I am VERY green when it comes to coding. I copied this code from another website that is working just fine but they are hooked up to a 4.3 version of mysql- can anyone help?This is the error-
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1, city, state, zip, ) VALUES ('', , ' 1', '', '', '','' )' at line 1
Here is my code:
mysql_select_db("membership70", $con);
$name=mysql_real_escape_string($_POST['Name']);
$address1=mysql_real_escape_string($_POST['Address1']);
$city=mysql_real_escape_string($_POST['city']);
$state=mysql_real_escape_string($_POST['state']);
$zip=mysql_real_escape_string($_POST['zip']);
$email=mysql_real_escape_string($_POST['email']);
$sql="INSERT INTO Members (name, email, adress 1, city, state, zip, ) VALUES ('$name', , '$address 1', '$city', '$state', '$zip','$email' )";
if (!mysql_query($sql,$con)) {
die('Error: ' . mysql_error());
}
You have speces in address [space] 1 both places.
$sql="INSERT INTO Members (name, email, adress 1, city, state, zip, ) VALUES ('$name', , '$address 1', '$city', '$state', '$zip','$email' )";
Also you have an extra colon after zip and the order of the values does not match the order you gave the column names. Try:
$sql="INSERT INTO Members (name, email, adress1, city, state, zip) VALUES ('$name', '$email', '$address1', '$city', '$state', '$zip')";
Get rid of the space in adress 1 and your variable $adress 1....not allowed (spaces that is)
That statement clearly tells you the SQL syntax is broken. Take a closer look into your statement in the $sql variable. echo it to debug it properly.
$sql = "INSERT INTO .....";
echo $sql; die();
At first glace, there shouldn't be a blank neither the column adress nor the corresponding php variable. And you should sort the order of your columns. email is declare second, so don't put the value of it at the end.
So, instead of...
$sql="INSERT INTO Members (name, email, adress 1, city, state, zip, ) VALUES ('$name', , '$address 1', '$city', '$state', '$zip','$email' )";
...try...
$sql="INSERT INTO Members (name, email, adress1, city, state, zip) VALUES ('$name', '$email', '$address1', '$city', '$state', '$zip' )";
You have a column adress 1 there, that's not a valid column name in mysql and it could not have worked on any mysql version. What columns does your table have?
Also, your second value is empty, put some value in there (or a default one). And a comma at the end of the columns list. Email at the wrong place.
Maybe this would work?
$sql = "INSERT INTO Members (name, email, adress1, city, state, zip) ".
"VALUES ('$name', '$email', '$address1', '$city', '$state', '$zip')";
(I'm not sure about the third column, should it spell address1?)