Insert records into a table

minds_gifts

hello everyone,
I'm trying to insert records into a subtopic table.First, I'm getting all the topics into a list box and then I'm selecting a topic and then will hit a button.It shows me a text box where i can enter a sub-topic name corresponding to the topic and will hit the insert button.
After the insert st is executed, it displays me the error mesg"Something is wrong in the syntax".
Could somebody pls figure out where the error is???
My subtopic table has 3 fields(TOPIC_ID, SUBTOPIC_ID, SUBTOPIC_NAME)SUBTOPIC_ID being auto-increment.

Thanks in advance

<?php
//Database connection

?>
<html>
<head>
<title>Insert subtopics</title>
</head>
<body>		   

<?
if (isset($HTTP_POST_VARS ["TOPIC_ID"])) {
    $TOPIC_ID=(int)$HTTP_POST_VARS ["TOPIC_ID"];
    $q=mysql_query("select * from topic where TOPIC_ID=$TOPIC_ID");
    if (!mysql_num_rows($q))
        $TOPIC_ID=0;
} else {
    $TOPIC_ID=0;
}

if ($TOPIC_ID) {
    $headline=trim($HTTP_POST_VARS ["headline"]);
    if (strlen($headline)) {
mysql_query("INSERT INTO subtopic (TOPIC_ID, SUBTOPIC_ID, SUBTOPIC_NAME) VALUES ($TOPIC_ID, $SUBTOPIC_ID, '$headline')")
or die("Invalid query: " . mysql_error()); 
        echo "Subtopic inserted!";
    }
}

// To display the form

echo "<html><form method=post action=\"$PHP_SELF\">";
echo "<select name=TOPIC_ID>";
$q=mysql_query("SELECT * FROM topic ORDER BY TOPIC_NAME");
while ($l=mysql_fetch_array($q)) {
     $selected="";
     if ($l["TOPIC_ID"]==$TOPIC_ID)
         $selected="selected=1";
     echo "<option value=\"".$l["TOPIC_ID"]."\" $selected>".$l["TOPIC_NAME"];
}
echo "</select>";

echo " <input type=submit name=submitname value=\"Change\"> ";
echo "<br><br>";
if ($TOPIC_ID) {
    echo "Headline: <input name=headline size=40><br>";
    echo "<input type=submit value=\"Insert\">";
}
echo "</form>";
?>
</body>
</html>

x3ro

try die ($sql) instead of die (mysql_error()) .. it might be something as obvious as addslashes() ..

trooper

Put all the vars in quotes like you have done with the '$headline'

I always use or die(mysql_error()) to help me debug my sql syntax. So put that at the end of your mysql_query line, if there is a problem then the mysql_error() will tell you what it is.

HTH

mysql_query("INSERT INTO subtopic ('TOPIC_ID', 'SUBTOPIC_ID', 'SUBTOPIC_NAME') VALUES ($TOPIC_ID, $SUBTOPIC_ID, '$headline')" or die(mysql_error())

lera

I think your query must be like that:

INSERT INTO subtopic
(TOPIC_ID, SUBTOPIC_ID, SUBTOPIC_NAME)
VALUES ('$TOPIC_ID', '', '$headline')

Because SUBTOPIC_ID is AutoIncrement. In your query $SUBTOPIC_ID every time = 0 (you have not field SUBTOPIC_ID in your form)

minds_gifts

Thanks everybody,

Though it does'nt show me any errors, I dont see the records getting inserted into the table.

Could somebody pls point out where the problem might be??

The field names are properly spelled with that of in the table, i only see there might be error with subtopic_id.Does i make sense??

Many thanks in advance

minds_gifts

well, now i see the records being inserted into the table.But Unfortunately, I see the SUBTOPIC_ID is zero with the record being inserted.There were 28 records before and now the SUBTOPIC_ID should be 29, but i see it as 0.The SUBTOPIC_ID is an auto-incrementing field in my table.why does'nt it takes the next id field when a new record is inserted.
Could you somebody pls fix that error

Heres the code:

<?php
//database connection
?>
<html>
<head>
<title>To insert subtopics</title>
</head>
<body>		   

<?
if (isset($HTTP_POST_VARS ["TOPIC_ID"])) {
    $TOPIC_ID=(int)$HTTP_POST_VARS ["TOPIC_ID"];
    $q=mysql_query("select * from topic where TOPIC_ID=$TOPIC_ID");
    if (!mysql_num_rows($q))
        $TOPIC_ID=0;
} else {
    $TOPIC_ID=0;
}

if ($TOPIC_ID) {
    $headline=trim($HTTP_POST_VARS ["headline"]);
    if (strlen($headline)) {
mysql_query("INSERT INTO subtopic (TOPIC_ID, SUBTOPIC_ID, SUBTOPIC_NAME) VALUES ('$TOPIC_ID', '$SUBTOPIC_ID', '$headline')")or die(mysql_error());
//or die("Invalid query: " . mysql_error()); 
        echo "Subtopic inserted!";
    }
}

// To display the form.

echo "<html><form method=post action=\"$PHP_SELF\">";
echo "<select name=TOPIC_ID>";
$q=mysql_query("SELECT * FROM topic ORDER BY TOPIC_NAME");
while ($l=mysql_fetch_array($q)) {
     $selected="";
     if ($l["TOPIC_ID"]==$TOPIC_ID)
         $selected="selected=1";
     echo "<option value=\"".$l["TOPIC_ID"]."\" $selected>".$l["TOPIC_NAME"];
}
echo "</select>";

echo " <input type=submit name=submitname value=\"Change\"> ";
echo "<br><br>";
if ($TOPIC_ID) {
    echo "Headline: <input name=headline size=40><br>";
    echo "<input type=submit value=\"Insert\">";
}
echo "</form>";
?>
</body>
</html>

minds_gifts

it was my mistake, I fixed it though.

Thanks everybody