I am getting; "Warning: oci_execute(): ORA-00904: "JAN": invalid identifier", when I try to execute these commands:
function stime($conn3, $time){
$result = oci_parse($conn3, "SELECT TO_CHAR($time, 'mm/dd/yyyy') FROM MON_EVENTS")or die(oci_error());
oci_execute($result);
}
STIME is also a date field in the database.
I am passing the STIME field to $time as stime($row_oci['STIME']).
You were bitten by PHP string interpolation:
$result = oci_parse($conn3, "SELECT TO_CHAR($time, 'mm/dd/yyyy') FROM MON_EVENTS")or die(oci_error());
// ^^^^^
$time is replaced by its content converted to a string -- and that before passing the value to the oci_parse function. As the string representation of a date might contain spaces, letters, /, ... it will confuse the Oracle SQL parser that report ORA-00904: Invalid identifier.
As of myself I would suggest to use bind parameter instead. This is much less error-prone -- and much more safe:
$result = oci_parse($conn3, "SELECT TO_CHAR(:time, 'mm/dd/yyyy') FROM MON_EVENTS");
oci_bind_by_name($result, ':time', $time);
$id = $row_oci['ID'];
$result = oci_parse($conn2, "SELECT TO_CHAR(STIME,'MON/DD/YY hh24:mm:ss') FROM MON_EVENTS WHERE ID = $id");
oci_execute($result);
while($row_result = oci_fetch_array($result)){
echo "". $row_result['0'] ."";}
Related
I have a DATE field in my table, and I'm trying to update it using the following code:
$query = mysqli_query($conn,$sql);
$todaydate = date("Y-m-d");
$sqlDate = date('Y-m-d', strtotime($todaydate));
$sql="UPDATE Library SET Loaned=1, LoanedDate=$sqlDate WHERE BookId=$bookId";
$query = mysqli_query($conn,$sql);
It updates the "Loaned" field fine, but always sets the Date field to "0000-00-00". Can anyone indicate what I'm doing wrong?
Youre missing ' quotes in your update query
Just use something like this
$sql="UPDATE Library SET Loaned=1, LoanedDate='$sqlDate' WHERE BookId='$bookId'";
$query = mysqli_query($conn,$sql);
You must have to add Quotes to the values... other wise the date may seem like an invalid integer to mysql. Here's how:
$query = mysqli_query($conn,$sql);
$todaydate = date("Y-m-d", time()); //<== DON'T FORGET THE 2ND ARGUMENT TO date(): TIME-STAMP. YOU MAY USE: time()
$sqlDate = date('Y-m-d', strtotime($todaydate));
$sql ="UPDATE Library SET Loaned=1, LoanedDate='{$sqlDate}' WHERE BookId='{$bookId}'";
$query = mysqli_query($conn,$sql);
Why don't you use only MySQL function to update date
Use this
$sql = "UPDATE Library SET Loaned = 1, LoanedDate = DATE(NOW())
WHERE BookId = '".$bookId."'";
Change your query with this code
$sql="UPDATE Library SET Loaned=1, LoanedDate= current_date() WHERE BookId='".$bookId."'";
Also check data structure in database.
This query is working fine:
$current_date = strtotime(date('Y-m-d H:i:s'));
$sql = "UPDATE user SET dt_added = '".$current_date."' WHERE id = '$id' ";
You're not concatenating the Date string properly in the SQL query.
Use ' and . operator to concatenate the string. Like this,
$sql="UPDATE Library SET Loaned=1, LoanedDate='".$sqlDate."' WHERE BookId='".$bookId."'";
Altough it's good practice to use Prepared statement to pass arguments in SQL query.
Learn more about PHP Prepared Statements
Looking closer, I would just do:
$sql="UPDATE Library SET Loaned=1, LoanedDate='".$todaydate."' WHERE BookId='".$bookId."'";
instead of
$sql="UPDATE Library SET Loaned=1, LoanedDate='".$sqlDate."' WHERE BookId=$bookId";
I am trying to build an SQL query that will insert the check-in time for a child at a fictional daycare facility. Here is a condensed version of my query code:
$childFirstName = $_POST['childFirstName'];
$childLastName = $_POST['childLastName'];
$now = new DateTime();
$nowDate = $now->format('m-d-Y');
$nowTime = $now->format('h:i');
$sql_childID = "SELECT id FROM child
WHERE firstName = '$childFirstName'
AND lastName = '$childLastName'";
$result = $pdo->query($sql_childID);
$row = $result->fetch();
$sql = "INSERT INTO checkinout(date, in, child_id) VALUES(?,?,?)";
$statement = $pdo->prepare($sql);
$statement->bindValue(1, $nowDate);
$statement->bindValue(2, $nowTime);
$statement->bindValue(3, $row['id']);
$statement->execute();
The checkinout table uses VARCHAR datatypes for the date and in columns. Originally they were set to use DATETIME, but I received the same errors.
Right now I get the following errors returned...
You can see from the error messages that my values are getting passed in the way I want them to, but I don't understand where my syntax error would be.
Enclose your field names with backticks. Two of them are reserved words (date and in):
$sql = "INSERT INTO checkinout(`date`, `in`, `child_id`) VALUES(?,?,?)";
https://dev.mysql.com/doc/refman/5.5/en/keywords.html
I've got the following PDO statement:
$myquery = SELECT * FROM DATABASE WHERE TABLE(Date) > 2014-04-07
$stmt = $this->pdo->prepare($myquery);
$stmt->execute();
I get the following error:
Operand type clash: date is incompatible with int
How is it possible to compare a Date in PDO?
Surround the date in single quotes when comparing:
$myquery = "SELECT * FROM DATABASE WHERE TABLE(Date) > '2014-04-07'";
I would try it like:
$myquery = SELECT * FROM table WHERE Date > ?
$stmt = $this->pdo->prepare($myquery,array('2014-04-07'));
$stmt->execute();
even better would be a non string for the date i.e. a datetime-object
I've got a PHP/MySQL script that is yielding strange results on a date field. All along the process, my dates are fine until the very end. The final result has every entry in the date field as '0000-00-00'. I'm totally stuck and don't know what else to do. I can tell that this is an issue with PHP not interpreting this as a date, but I don't know how to fix it. Here is my code:
$sql = "CREATE TABLE temp_workouts (my_date date, sg_id int(11), loc_id int(11))";
$result = mysql_query($sql);
if (!$result) {
$tag_success = "failure";
$tag_message = mysql_error();
echo encodeJSON($tag_success, $tag_message);
die();
}
$sql = "SELECT * FROM my_table";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$my_date = $row['my_date'];
echo $my_date . " "; //<--this output looks perfect
$sql = "INSERT INTO temp_table (my_date) VALUES ($my_date)";
$result2 = mysql_query($sql);
}
die();
When I flip over to MyPHPAdmin and look at the table, the entire column my_date contains '0000-00-00'. How can I get PHP to recognize this as a 'Y-m-d' formatted date? Thanks. I appreciate any help.
I suspect the issue is that you haven't enclosed a string literal in single quotes:
INSERT INTO temp_table (my_date) VALUES ('$my_date')
^--- ^--- string literals in single quotes
Otherwise, the statement is probably something like:
... VALUES (2013-08-22)
MySQL isn't converting that into a valid date, issuing a warning message, and inserting a "zero" date.
Your immediate problem is that you don't use quotes around date values in your insert statement.
Change
$sql = "INSERT INTO temp_table (my_date) VALUES ($my_date)";
to
$sql = "INSERT INTO temp_table (my_date) VALUES ('$my_date')";
^ ^
Now, you can just use INSERT ... SELECT syntax to achieve your goal in one go
INSERT INTO temp_table (my_date)
SELECT my_date
FROM my_table
Therefore this part of your code
$sql = "SELECT * FROM my_table";
$result = mysql_query($sql);
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$my_date = $row['my_date'];
echo $my_date . " "; //<--this output looks perfect
$sql = "INSERT INTO temp_table (my_date) VALUES ($my_date)";
$result2 = mysql_query($sql);
}
can be changed to
$sql = "INSERT INTO temp_table (my_date)
SELECT my_date FROM my_table";
$result2 = mysql_query($sql);
On a side note: Consider switching to either PDO or MySQLi and use prepared statements.
Try this one...This will re-convert it to date, and then save..
$dt = strtotime($row['my_date']);
$date = date("Y-m-d",$dt);
$sql = "INSERT INTO temp_table (my_date) VALUES ({$date})";
I have this code:
public function updateOrder($num, $ufood, $uquan) {
$response = array();
mysql_query("SET NAMES 'utf8'");
foreach ($ufood as $index => $f) {
$result = mysql_query("SELECT food, quantity, uquantity FROM table1 WHERE food ='".$f."'") or die(mysql_error());
$no_of_rows = mysql_num_rows($result);
$response['number rows'] = $no_of_rows;
if ($no_of_rows>0) {
while ($row = mysqli_fetch_array($result)); {
if (!$row['uquantity']) {
$w = "INSERT INTO table1(uquantity) VALUES ('$uquan[$index]')";
mysql_query($w);
$e = (int)$row['quantity'];
$q = (int)$uquan[$index];
$sum = $e+$q;
$s = (string)$sum;
$d = "UPDATE table1 SET quantity = '$s' WHERE food = ".$row['$food']." ";
mysql_query($d);
} else if($row['uquantity']) {
$c = (int)$row['uquantity'];
$q = (int)$uquan[$index];
$sumq = $c+$q;
$sq = (string)$sumq;
$d = "UPDATE table1 SET uquantity = '$sq' WHERE food = ".$row['$food']." ";
}
}
} else {
$string ="INSERT INTO table1(food,uquantity) VALUES ('".$f."','".$uquan[$index]."')";
$z = mysql_query($string);
}
}
}
Well i can not make this work, and i am trying all kinds of things put still it doesn't work.
So i have some questions:
Is this structure of foreach and while valid?
Though the $result query returns some rows from the database, when i try to use $row['quantity'], as a value, i get null.
In this code i receive some data from an android app, and i try to "see", if there are already entries for the type food of my db_table(table1). If there are entries i want the db to sum the quantity entry of the android sent, data with the one that are inside my db, and update the field. This is the basically it. But as i said when i try to use the data that comes from the database, i get null values.
Please if someone could give me some hint, cause I'm really stuck..
There are many problems with your code. I'm marking this answer as Community Wiki, and I invite others to edit and add things as they find them.
You may also consider posting to https://codereview.stackexchange.com/ instead, when you have so many mistakes, until you have a more specific question.
Bad variable interpolation
This line won't do what you want it to:
$w = "INSERT INTO table1(uquantity) VALUES ('$uquan[$index]')";
This is not quite valid PHP syntax. You can either concatenate expressions:
$w = "INSERT INTO table1(uquantity) VALUES ('".$uquan[$index]."')";
Or you can embed expressions in curly braces:
$w = "INSERT INTO table1(uquantity) VALUES ('{$uquan[$index]}')";
Or you can use a query parameter placeholder:
$w = "INSERT INTO table1(uquantity) VALUES (?)";
$stmt = mysqli_prepare($w) or die(mysqli_error());
$uqi = $uquan[$index];
mysqli_stmt_bind_param($stmt, "i", $uqi);
mysqli_stmt_execute($stmt);
Mixing MySQL APIs
You can't mix mysql_query() with mysqli_fetch_array(). PHP has more than one API for MySQL, and you can't mix them. You should standardize on using the mysqli API, because the older mysql API is now deprecated.
Semicolon defeats while loop
The semicolon after the while statement makes the loop a no-op, and when it terminates, the $row contains nothing.
while ($row = mysqli_fetch_array($result)); {
Should be:
while ($row = mysqli_fetch_array($result)) {
Using variables inappropriately
Referencing a $row key with a single-quoted variable is probably not what you mean, in multiple ways:
$d = "UPDATE table1 SET quantity = '$s' WHERE food = ".$row['$food']." ";
The column name in the select-list of your earlier SELECT query is 'food', not '$food'.
Also, even if you meant to use a variable name $food as the key, putting it in single quotes would not use the value of the variable, it would be the literal string '$food'.
Failure to quote string literal?
Furthermore, you use a quoted literal for comparing to the food column in your SELECT query, which makes me think it might be a string.
So the UPDATE should be something like:
$d = "UPDATE table1 SET quantity = '$s' WHERE food = '".$row['food']."' ";
Or:
$d = "UPDATE table1 SET quantity = '$s' WHERE food = " . intval($row['food']);
Or preferably use parameters and a prepared query, then you don't need to worry about quotes or types:
$d = "UPDATE table1 SET quantity = ? WHERE food = ?";
. . .
Failure to check for errors
Every query might fail, either because you have a syntax error (e.g. a string without quoting), or because the table doesn't have a column by the name you reference, or privileges issues, etc.
Always check the return status of the query function when you run a SQL query. The function returns false if there's an error, and if that happens you must check the error message.
mysqli_query($mysqli, $d) or trigger_error(mysqli_error($mysqli), E_USER_ERROR);
Failure to execute the UPDATE
Your second update assigns a SQL query string to the variable $d, but then does not execute that update query at all!