I wrote some PHP code to retrieve some data from my MySQL database.
My NetBeans debugger (v7.4) shows all array elements and related keys (of $data_array_from_db) which keys corresponds with the column fields in a the database table, EXCEPT the keys of which database fields are not filled (value NULL).
However, when the array_key_exists() function is executed in the code below for say 'akey' that corresponds with such non filled database field, array_key_exists() returns a 'true' value (instead of expected false) - as if the key does exist (while again, the NetBeans debugger does not show the $data_array_from_db['akey']).
I know for sure that the database function array_key_exists() works correctly. Am I interpreting something wrong? Does/should the key exist if its corresponding database value is NULL?
$data_array_from_db = $corpdb->GetSpecificDBRecords($sqlquery5); //GetSpecificDBRecords() includes some PDO statements
if (array_key_exists('akey', $data_array_from_db)) { // Database value is optional; exists?
$response_array[0]['akey'] = $data_array_from_db['akey']; // This line executed while akey does not show in NetBeans debuggers
}
else { // Database value does not exist; set to 0
$response_array[0]['akey'] = 0;
}
NULL is a value, and you need to check for it if you want to replace it. Change:
if(array_key_exists('akey',$data_array_from_db) ){
To:
if(array_key_exists('akey',$data_array_from_db) && !is_null($data_array_from_db['akey'])){
Or just check if it exists and it's not empty:
if (!empty($data_array_from_db['akey'])) {
Related
I have a simple MySQLi statement with bound results. One of the fields in that recordset is a double and results can be either NULL, 0 or other numeric value.
$wheels_cost is the recordset bound result.
The following logic should ignore null values and only process zero or other numeric values.
if($wheels_cost <> NULL){
carry out a process
} else {
the value is null so ignore
}
For some reason, the results I'm getting seem to reflect that the result is a zero rather than a null - which is not the desired result.
I've checked the DB and the field is null.
I need to tell the difference between a null and a zero value.
I've only tried the above logic and can't think why it doesn't work. Should I be using isset() or is_null() or empty()?
Thanks for feedback
Finished code below:
if(!is_null($wheels_cost)){
carry out a process
} else {
the value is null so ignore
}
Worked perfectly
Try this:
if($wheels_cost is not NULL){
# carry out a process
} else {
# the value is null so ignore
}
I do not use mySQLi but I think the syntax should look like PostGreSQL
In PostGreSQL, the NULL value can not be equal or differentiated to any column or variable, since it is an unknown value and of an unknown type.
Examples here: https://www.techonthenet.com/mysql/is_null.php
I have a table with the column data type like this:
price decimal(6,2) unsigned NOT NULL,
And my validation code is like this:
if ( (!empty($_POST['price'])) && (is_numeric($_POST['price'])) && (isset($_POST['price'])) > 0 ) {
$price = ($_POST['price']);
} else{
$price = FALSE;
echo '<p> Only accept number and must be higher than zero</p>';
}
I use the $_POST form for users to submit the value. But you know,
1/ When the user types any non-numeric value such as a,b,c etc, it also validates well.
2/ When the user types the value zero in, it validates well too.
However, the question is that when I tested it with no value typed in. I mean that I left the value empty and hit the 'submit' button, the error message still returned as per the }else { part does, but the value was still inserted into the table column with a value of 0.00 technically at the same time.
To my limited knowledge, I can guess that the problem was probably at the Mysqli data type of the table I chose, but i don't know how to correct it.
Can you help me, please?
I think the solution you're looking for is to simply move the inserting code to the first if statement. That way it'll only insert the value if it is numeric and not empty.
What you describe means that you've failed to stop the insert when $price===false
(i.e. the problem is not the evaluation; it has given you the correct message. You've some programming logic error elsewhere)
I have to retrieve the history of a user and I have 4 tables whose data depend on each other.I can retrieve the data using loops,but I instead used the "where IN ()" clause and I implode the output of the previous query.However,if the list I provide to "where IN()" is empty it return an error.Is it that IN() cannot be empty?
When imploding an array for the IN clause, i do one of two things
1: Check if you even need to run the query at all
if(!empty($some_array)) {
//run mysql query
}
else {
// if you need to do something if the array is empty, such as error or set some defaults, do it here
}
2: A value in the array initiliser which is not ever in the database (for example, if im selecting based on a auto incrememnt id, i use zero as a default array value to stop any issues with empty data sets, as zero will never be in my id column).
$some_array = array(0);
You can add an empty value to the start, such as IN (0,your values here)
$blanknumber = $_POST["blankstartnumber"];
while ($blanknumber <= ($_POST["blankendnumber"] ))
{
echo "$blanknumber";
$blankid = $blanknumber;
$query = "INSERT INTO blank (Blank_ID) VALUES ('$blankid')";
mysql_query($query,$con);
$blanknumber++;
}
So the values are added into the database. Lets say if I have the starting number at 1 and ending at 5. It will all the those values, but it's still trying to add more into the database. I also tried adding an IF statement aswell. if ($blanknumber != $_POST["blankendnumber"])
12345 Error: Duplicate entry '5' for
key 'PRIMARY'
Make sure your $POST value is an integer; by default, I believe it will be cast as a string.
$_POST['varName'] = (int) $_POST['varName'];
edit:
$blanknumber = $_POST["blankstartnumber"];
while ($blanknumber <= ($_POST["blankendnumber"] ))
This should only execute once, since you're setting both comparison variables equal. Definitely 2x check your code.
The database error indicates that Blank_ID is your primary key for that table, and you'd already inserted a 5 into the row. A primary key's values can exist only once in the entire table - duplicates are forbidden (if they were allowed, it wouldn't be a primary key anymore).
If your while loop isn't ending, I'd suggest dumping out both the blankendnumber and blankstartnumber before the loop starts, making sure you've got the right values in there.
It looks like it's actually functioning properly, but you might not have tidy'ed up your db table prior to running. If your output was:
123455 Error: Duplicate entry '5'...
Then, you'd have a programming error, as 5 is getting run twice. Instead, I think you already have data in the blank table that causes a conflict.
Edit: to automatically have MySQL handle the duplicate key error gracefully, you can use the ON DUPLICATE KEY clause to update the row.
INSERT INTO blank (Blank_ID) VALUES (5) ON DUPLICATE KEY UPDATE mod_date = NOW();
I have a MYSQL table with an ENUM field named "offset" and some other columns. The field is defined as:
ENUM(0,1), can be NULL, predefined value NULL
Now I have two server. A production server and a development server and the same PHP script used to create and to update the database.
First step: the application create the record witout passing the "offset" in the CREATE query.
Second step: the application ask to the user some data (not the "offset" value), read the row inserted in step one and make an array, update some field (not the "offset" field), create a query in an automated fashion and save the row again with the updated values.
The automated query builder simple read all the field passed in an array and create the UPDATE string.
In both systems I obtain this array:
$values = array(... 'offset' => null);
and convert it in this same query passing the values in the mysql_real_escape_string:
UPDATE MyTable SET values..., `offset` = '' WHERE id = '10';
Now there is the problem. When i launch the query in the production system, the row is saved, in the development system I got an error and the db says that the offset data is wrong without saving the row.
From phpmyadmin when I create the row with the first step, it shows NULL in the offset field. After saving the field in the system which give no errors, it show me an empty string.
Both system are using MySQL 5 but the production uses 5.0.51 on Linux and development use 5.0.37 on Windows.
The questions:
Why one system give me an error an the other one save the field ? Is a configuration difference ?
Why when I save the field which is an enum "0" or "1" it saves "" and not NULL ?
Why one system give me an error an the other one save the field ? Is a configuration difference ?
Probably. See below.
Why when I save the field which is an enum "0" or "1" it saves "" and not NULL ?
According to the MySQL ENUM documentation:
The value may also be the empty string ('') or NULL under certain circumstances:
If you insert an invalid value into an ENUM (that is, a string not present in the list of permitted values), the empty string is inserted instead as a special error value. This string can be distinguished from a "normal" empty string by the fact that this string has the numeric value 0. ...
If strict SQL mode is enabled, attempts to insert invalid ENUM values result in an error.
(Emphasis added.)
strager's answer seems like a good explanation on why your code behaves differently on the 2 environments.
The problem lies elsewhere though. If you want to set a value to NULL in the query you shound use exactly NULL, but you are using mysql_real_escape_string() which result is always a string:
$ php -r 'var_dump(mysql_real_escape_string(null));'
string(0) ""
You should handle this differently. E.g:
$value = null
$escaped_value = is_null($value) ? "NULL" : mysql_real_escape_string($value);
var_dump($escaped_value);
// NULL
Some DB layers, like PDO, handle this just fine for you.
If you want it to be NULL, why don't you do this in the first place:
UPDATE MyTable SET values..., `offset` = NULL WHERE id = 10;