inserting data into a databse...twice?

[Henri]

whenever I run my script to update news, it inserts the "stuff (text)" to the database, but does it twice, thus resulting in having a duplicate of every news item I post. my code is the following:

$sql = "INSERT INTO SEA_news (content) VALUES ('$add_news')";
mysql_query("$sql", $connect) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());

I really don't see why it inserts the text twice. any help would be greatly appreciated🙂

[hand]

This insert statement inserts only ONE record to db. (An insert is not an update)

[theslinky]

let us see some more of the code. are you sure you're not calling mysql_query twice?

-sridhar

[Henri]

I would'nt think so...well, here's the full code:

$add_news = "<table width=\"100%\" cellpadding=\"0\" cellspacing=\"0\" border=\"0\">
<TR height=\"30\">
<TD bgcolor=\"#000000\" width=\"450\" align=\"left\" valign=\"middle\"><font size=\"2\"><b>$article_subject</b></font> - <font size=\"2\">Posted @ $date by $username</font></TD></TR>
<TR>
<TD bgcolor=\"#002660\" width=\"450\" align=\"left\" valign=\"top\"><p>$article_content</TD></TR>
</table>
<br><br>";
$sql = "INSERT INTO SEA_news (content) VALUES ('$add_news')";
mysql_query("$sql", $connect) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
$query = mysql_query("$sql", $connect);
if ($query == TRUE) {
echo "<center>News was successfully updated.</center>";
}else{
echo "<center>There was a problem when updating the news, please try again.</center>";

[dc2_jspec]

looks to me like mysql_query is being called here twice.

mysql_query("$sql", $connect) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
$query = mysql_query("$sql", $connect);

[theslinky]

Henri,

you ARE calling the mysql_query TWICE:

$sql = "INSERT INTO SEA_news (content) VALUES ('$add_news')";
mysql_query("$sql", $connect) or die("Database choked on query ($sql). MySQL sez: ".mysql_error());
$query = mysql_query("$sql", $connect);

If you want to trap errors, do this:

$sql = "INSERT INTO SEA_news (content) VALUES ('$add_news')";
$query = mysql_query("$sql", $connect);
if($query) {
    // it executed
}
else {
    die("Database choked on query ($sql). MySQL sez: " . mysql_error());
}

-sridhar

[Henri]

so just by setting a mysql_query("$query", $connect); to a variable is gets executed?

[theslinky]

of course! i mean, how else do you think $query will get a value set unless the PHP executes the right hand side of the = sign?

so that's why i suggest your run only $query = mysql_query() and check whether $query is false. if it's false, then use mysql_error to find the error. otherwise you know it executed.

-sridhar

[Henri]

ah, okay, thanx a lot for your help. but now, my code to fetch it does'nt show the news🙁

EDIT: very weird....it displays all news except for the first entry. is there any way to fix that?

<?php
$db_name = "novaz3ro_com"; 
$connect = mysql_connect('localhost','***','***'); 
$db = mysql_select_db($db_name,$connect);

$sql = mysql_query("SELECT content FROM SEA_news", $connect);
$news = mysql_fetch_row($sql) or die("No news to be displayed.");
while($row = mysql_fetch_array($sql)) {
echo "$row[content]<br><br>";
}
?>