mysql_query不返回资源

Question

I've been all up and down StackExchange, and much of the internet's cornucopia of lesser sites, looking for any good reason my code doesn't work, but this one has me stumped. I'd like to display the total number of rows in a particular MySQL table, and set the count as a variable, to use later in the script. When I run the following, the script dies, and I get a PHP warning, telling me that mysql_result() expects parameter 1 to be resource, string given.

$conn = mysql_connect('mysql_server', 'username', 'password');
if (!$conn) {
    die('Connect Error ' . mysql_error());
}
mysql_select_db('my_database', $conn);
if (!mysql_select_db('my_database')) {
    die('Could not select database: ' . mysql_error());
}
$max_count_query = ("SELECT COUNT(*) FROM table");
// Perform Query
$max_count_action = mysql_query($max_count_query);
$mcount = mysql_result($max_count_action, 0, 0);
printf("\nNumber of Records to Process: ", $mcount);

What does the collective genius of StackOverflow think?

In response to the comments, I have another mini-slab of code to offer:

$conn = mysql_connect('127.0.0.1', 'username', 'password');
if (!$conn) {
    die('Connect Error ' . mysql_error());
}
mysql_select_db('my_database', $conn);
if (!mysql_select_db('my_database')) {
    die('Could not select database: ' . mysql_error());
}
$max_count_query = ("SELECT COUNT(*) FROM directory_nap");
// Perform Query
$max_count_action = mysql_query($max_count_query);
if (!$max_count_action){
    die('mysql query error: ' . mysql_error());
    }
$mcount = mysql_result($max_count_action, 0, 0);
if (!$mcount){
    die('mysql result error: ' . mysql_error());
    }
printf("\nNumber of Records to Process: ", $mcount);

Answer 1

Provided the table tablename is just a sample (see answer by bitfox), there's nothing wrong with your code. I can use the same code on my test server and get results by changing the table name to one that I know exists in my own db.

What is most troubling, however, is that you indicate the error says mysql_result() expects parameter 1 to be resource, string given -- if your SQL has an error, mysql_query returns boolean ( docs ), not a string. So, you're either not showing the same test code as you're actually using, or you've given us an inaccurate error message.

At some point, you must be assigning a string into the variable $max_count_action . Here's what I get when I send a query with an intentional problem: Warning: mysql_result() expects parameter 1 to be resource, boolean given -- note "boolean", not string.

So, I think your first step is to choose a different table name. That said, if you're using a reserved word as a table or column name you can still access it by surrounding the string in the backtick (`) character:

SELECT COUNT(*) FROM `table`

Second step is to see what's happening to $max_count_action to turn it into a string. Finally, use mysql_error consistently to debug, I would suggest doing something a little nicer with it than die for production code, however.

// working code on my test server
$max_count_query = ("SELECT COUNT(*) FROM users");
$max_count_action = mysql_query($max_count_query);
$mcount = $max_count_action ? mysql_result($max_count_action, 0) : 'Error: '.mysql_error();
print "\nNumber of Records to Process: ". $mcount;

Answer 2

it's very likely that the error is here:

SELECT COUNT(*) FROM table

" table " is a reserved word of SQL language. You should change it.

Answer 3

The cause of the problem is not clear from the code sample (it could be a wrong table name, the parentheses around the query string, and so on), however there are a number of inaccuracies in it. You can try the following code, it should at least fix a couple side bugs and give you more details on the error:

$conn = mysql_connect('mysql_server', 'username', 'password');
if ($conn === false) {
    die('Connect Error ' . mysql_error($conn));
}
if (mysql_select_db('my_database', $conn) === false) {
    die('Could not select database: ' . mysql_error($conn));
}
$max_count_query = "SELECT COUNT(*) FROM table";
// Perform Query
$max_count_action = mysql_query($max_count_query, $conn);
if($max_count_action === false) {
    die('Query error ' . mysql_error($conn));
}
$mcount = mysql_result($max_count_action, 0, 0);
if($mcount === false) {
    die('Result retrieval error ' . mysql_error($conn));
}
printf("\nNumber of Records to Process: %s", $mcount);
mysql_free_result($max_count_action);

Hope this helps, bye!

Answer 4

Is the 'table' in your question a placeholder or a real name in your code? Try to change another name of your 'table' table, otherwise I would like to try mysql_fetch_array instead of mysql_result

$max_count_query = "SELECT COUNT(*) FROM t";
// Perform Query
$max_count_action = mysql_query($max_count_query) or die (mysql_error());
$mcount = mysql_fetch_array($max_count_action);
printf("\nNumber of Records to Process: ", $mcount[0]);

Answer 5

$conn = mysql_connect('mysql_server', 'username', 'password');
if (!$conn) {
    die('Connect Error ' . mysql_error());
}

This is OK, but the connect and coditional don't need to be split across 2 lines

mysql_select_db('my_database', $conn);

Again, this is questionable. If you're using multiple databases then referencing databases in your SQL is much simpler and safer than tracking state within your PHP code.

if (!mysql_select_db('my_database')) {
   die('Could not select database: ' . mysql_error());
}

Haven't you already done that?

$max_count_query = ("SELECT COUNT(*) FROM table");

Why the brackets?

// Perform Query

Isn't that obvious from the code?

$max_count_action = mysql_query($max_count_query);
$mcount = mysql_result($max_count_action, 0, 0);

It would be neater to pass the db handle to the mysql_query call.

You checked for an error after connecting, you checked for an error after switching database, but you don't check for an error after mysql_query() ?

printf("\nNumber of Records to Process: ", $mcount);

There is no placeholder in the format string for the mcount argument.Try:

printf("\nNumber of Records to Process: %d", $mcount);

I get a PHP warning, telling me that mysql_result() expects parameter 1 to be resource, string given.

Then the code you are running is not the code you've shown us; mysql_query() will not return a string.