MYSQL语法错误

Question

HI everyone i tried for 3 days and i'm not able to solve this problem. This is the codes and i have went through it again and again but i found no errors. I tried at a blank page and it worked but when i put it inside the calendar it has the syntax error. Thanks a million for whoever who can assist.

/** QUERY THE DATABASE FOR AN ENTRY FOR THIS DAY !!  IF MATCHES FOUND, PRINT THEM !! **/
$testquery = mysql_query("SELECT orgid FROM sub WHERE userid='$userid'");
while($row4 = mysql_fetch_assoc($testquery))
{
  $org = $row4['orgid'];
  echo "$org<br>";
  $test2 = mysql_query("SELECT nameevent FROM event WHERE `userid`=$org AND EXTRACT(YEAR FROM startdate)='2010' AND EXTRACT(MONTH FROM startdate)='08' AND EXTRACT(DAY FROM startdate)='15'") or die(mysql_error());
  while($row5=mysql_fetch_assoc($test2))
  {
    $namethis = $row5['nameevent'];
    $calendar.=$namethis;
  }
}

Answer 1

First question: what calendar are you talking about?

And here are my 2-cents: does the EXTRACT function returns a string or a number? Are the "backticks" ( userid ) really in your query? Try to strip them off. Bye!

Answer 2

It's a guess, given that you haven't provided the error message you're seeing, but I imagine that userid is a text field and so the value $org in the WHERE clause needs quotes around it. I say this as the commented out testquery has quotes around the userid field, although I appreciate that it works on a different table. Anyway try this:

SELECT nameevent FROM event WHERE userid = '$org' AND EXTRACT(YEAR FROM startdate)='2010' AND EXTRACT(MONTH FROM startdate)='08' AND EXTRACT(DAY FROM startdate)='15'

In such cases it's often useful to echo the sql statement and run it using a database client

Answer 3

First step in debugging problems like this, is to print out the acutal statement you are running. I don't know PHP, but can you first build up the SQL and then print it before calling mysql_query()?

EXTRACT() returns a number not a character value, so you don't need the single quotes when comparing EXTRACT(YEAR FROM startdate) = 2010 , but I doubt that this would throw an error (unlike in other databases) but there might be a system configuration that does this.

Another thing that looks a bit strange by just looking at the names of your columns/variables: you are first retrieving a column orgid from the user table. But you compare that to the userid column in the event table. Shouldn't you also be using $userid to retrieve from the event table?

Also in the first query you are putting single quotes around $userid while you are not doing that for the userid column in the event table. Is userid a number or a string? Numbers don't need single quotes.

Answer 4

Any of the mysql_* functions can fail. You have to test all the return values and if one of them indicates an error (usually when the function returns false ) your script has to handle it somehow.

---

Eg in your query

mysql_query("SELECT orgid FROM sub WHERE userid='$userid'")

you mix a parameter into the sql statement. Have you assured that this value (the value of $userid) is secure for this purpose? see http://en.wikipedia.org/wiki/SQL_injection

---

You can use a JOIN statement two combine your two sql queryies into one.

---

see also:

• http://docs.php.net/mysql_error
• http://docs.php.net/mysql_real_escape_string
• http://www.w3schools.com/sql/sql_join.asp

  ---

Example of rudimentary error handling:

$mysql = mysql_connect('Fill in', 'the correct', 'values here');
if ( !$mysql ) { // some went wrong, error hanlding here
  echo 'connection failed. ', mysql_error();
  return;
}

$result = mysql_select_db('dbname', $mysql);
if (!$result ) {
  echo 'select_db failed. ', mysql_error($mysql);
  return;
}

// Is it safe to use $userid as a parmeter within an sql statement?
// see http://docs.php.net/mysql_real_escape_string
$sql = "SELECT orgid FROM sub WHERE userid='$userid'";
$testquery  = mysql_query($sql, $mysql);
if (!$testquery  ) {
  echo 'query failed. ', mysql_error($mysql), "<br />\n";
  echo 'query=<pre>', $sql, '</pre>';
  return;
}