Passing Variable from PHP script to MYSQLI query

Question

I am trying to pass a variable to a mysqli query as shown below but it never updates the record

This works:

mysql> UPDATE zeus.alldata SET A='56.1' WHERE wwxRecord = 
(SELECT max(dateTime) FROM wwx.wmx_archive);
Query OK, 1 row affected (0.17 sec)
Rows matched: 1  Changed: 1  Warnings: 0

This does not:

$A = "56.1";
$sql = mysqli_query($conn,"UPDATE ".$template_db.".alldata SET A='$A' 
WHERE wwxRecord = (SELECT max(dateTime) FROM ".$wwx_db.".wmx_archive");

This does not:

$A = "56.1";
$sql = mysqli_query($conn,"UPDATE ".$template_db.".alldata SET A=".$A." 
WHERE wwxRecord = (SELECT max(dateTime) FROM ".$wwx_db.".wmx_archive");

Using an echo $A; right before and after the statement show the correct 56.1, but nothing I seem to try allows me to pass the $A to the query. The other variables work fine (".$template_db.") as I use them throughout the script for other database updates.

I hope I am just missing something simple!

if ( !$sql ) trigger_error( mysqli_errno($conn) . ' '. mysqli_error($conn), E_USER_ERROR); What does that print? — VolkerK
– VolkerK, Commented Mar 30, 2016 at 16:48
Are you executing all queries on the same system or 1st on local and other one on live? — itzmukeshy7
– itzmukeshy7, Commented Mar 30, 2016 at 16:54

A l w a y s S u n n y · Accepted Answer · 2016-03-30 16:58:22Z

1

You are missing the ending ) of mysqli_query() function

$A = "56.1";
$template_db = "zeus";
$wwx_db = "wwx";

$sql_string = "UPDATE ".$template_db.".alldata SET A='$A' 
WHERE wwxRecord = (SELECT max(dateTime) FROM ".$wwx_db.".wmx_archive)";

$sql_query = mysqli_query($conn,$sql_string);

edited Mar 30, 2016 at 16:58

answered Mar 30, 2016 at 16:52

A l w a y s S u n n y

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1 Comment

Richard Over a year ago

Wow...amazing...I must have looked at that thing for an hour and missed the missing ')'. Thank you!! It is working perfectly now!

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