Comparing variable value with result of a query in PHP [duplicate] - php

Question

This question already has answers here:
mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource
(31 answers)
Closed 8 years ago.
I'm trying to pull all the "totalsDate" values from a table "totals" and then either INSERT a new record or UPDATE an existing one based on a variable $date.
//getting all the dates from the totals table and assigning to row
$sqlDate = "SELECT totalsDate FROM totals";
$query = mysqli_query($dbCon, $sqlDate);
//$row = mysqli_fetch_array($query);
while($row = mysqli_fetch_array($query)){
$rowDate = $row['totalsDate'];
//if statement to either update the totals table or create a new record
if($rowDate = $date){
$sqlThree = "UPDATE totals SET lodgements = '$lodgementsAfter' WHERE branch_name = '$branchTest' AND totalsDate = '$date'";
$query = mysqli_query($dbCon, $sqlThree);
}
else {
$sqlFour = "INSERT INTO totals VALUES(NULL, '$branchTest', 0, '$amount', 0, '$date')";
$query = mysqli_query($dbCon, $sqlFour);
}
}
The update part works, however my else statement will never be executed and a new record cannot be entered. I also get this error:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in
I'm a bit confused on using the mysqli_fetch_array and how to actually use the data?
Thanks for any help.

Answer 1

One of the problem is that $query gets overwritten inside the loop; see here:
while($row = mysqli_fetch_array($query)){
// ...
$query = mysqli_query($dbCon, $sqlThree);
// ...
}
You should either a) don't store the result of mysqli_query() at all, or b) choose a different variable name, e.g. $update_res = mysqli_query(...);.
Better yet, use a single query to do both:
INSERT INTO totals VALUES (NULL, :branch, 0, :amount, 0, :date)
ON DUPLICATE KEY UPDATE lodgements = :lodgements
Just make sure the proper unique constraints are defined on the table.

Answer 2

your db query is not correct. You are not getting a good result back from your query. The error is saying it expects the result to be an array and its not. its returning a true of false.
read this mysqli_query
for more help.
do you have access to the db command line? what does the query return. Are you certain your db connection info is correct?
i found it.
$rowDate = $date should be $rowDate == $date or $rowDate === $date
your sql result is an array
$row[0] = "1st result"
$row[1] = "2nd result"
..etc
so theortically if you have more then 1 result, it should be doing multiple updates/inserts. are you seeing that? if its just one result then clearly 2014-02-13, 2014-02-26 are not the same.
$sqlDate = "SELECT totalsDate FROM totals where totalsDate='" . $date . "'";
this will return results, with only records that have $date.
$rowcount=mysqli_num_rows($query);
if ($rowcount) < 1))
{
insert;
}
else
{
update;
}