开发者

Problem with 'mysql_fetch_array() supplied argument is not a valid MySQL result resource'

开发者 https://www.devze.com 2023-01-26 15:18 出处:网络
I have a database with a table which includes among others, one column of dates (the column is titled \'date\') in the format of Y-m-d.

I have a database with a table which includes among others, one column of dates (the column is titled 'date') in the format of Y-m-d.

I want to extract all the rows of data which have been logged during the current month, so I am using the following query:

$year_month = date("Y-m");
$query = "SELECT date FROM tracker WHERE date LIKE '$year_month%'";

When I execute the query and attempt to output it visually, I get the error "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource on line 64".

Line 64 is the beginning of the following loop:

while($row = mysql_fetch_array($result, MYSQL_ASSOC)
{
    echo "Name :{$row['date']} <br>";
}

I've scoured google and had a look here too, but I can't seem to figure out this seemingly simple problem. As far as I can tell, the actual query is executing just fine.

开发者_开发问答

Any ideas?

** Additional - Call to MySQL Query Below **

$result = mysqli_query($conn, $query);

$conn refers to the following line:

$conn = @mysqli_connect($dbhost, $dbuser, $dbpass, $dbname) or die ('Error connecting to MySQL');


Two things jump out at me:

  1. Are you sure that's the query that goes with that result? Queries like insert and delete do not have results. They return a boolean indicating success.

  2. Are you checking to make sure the query succeeded? After the line where you call mysql_query, make sure you check the result. If it's false, you can call mysql_get_error to find out what went wrong.

EDIT: Okay, your real problem is that you're mixing mysql_* and mysqli_* commands. You need to use one or the other.


Something like this should work nicely...

if(!$result)
  throw new Exception("MySQL Error: ".mysql_error()." (#".mysql_errorno().")");

if(mysql_num_rows($result) <= 0)
  echo "No results found.";
else
  while($row = mysql_fetch_array($result, MYSQL_ASSOC))
    echo "Name :{$row['date']} <br>";

If you don't understand the error that you're getting from mysql_error() and mysql_errorno(), then post the output and we should be able to help with it.


$query = "SELECT date FROM tracker WHERE date LIKE '$year_month%'"; 

should be

$query = "SELECT `date` FROM tracker WHERE `date` LIKE '$year_month%'"; 

date is a reserved word in sql.

look for more examples here


What is the data type for the date field, if it is date type, then you should try following :

$query = "SELECT date FROM tracker WHERE year(date)='".date("Y")." and month(date) = '".date("m")."'";
$year_month = date("Y-m");

$result = mysql_query($query);
while($row = mysql_fetch_array($result, MYSQL_ASSOC)
{
    echo "Name :{$row['date']} <br>";
}
0

精彩评论

暂无评论...
验证码 换一张
取 消