Mysql Insert Statement from PHP

Question

I am trying to run PHP could which inserts a value into a MySQL table.

My db connection is working ok.

I have the code in a function:

function InsertRighmoveID($RightmoveID)
{
    # Connection already created in main program

    #Define the query to insert righmove ID already in Database
    #VALUES ('" . substr($name, 23, 31) . "')";

    echo "In InsertRighmoveID() function </br>";
        $query_enter_rightmove_ID = 

        "INSERT INTO tblRightMoveIDs (rightmoveID)
        VALUES ('" . $RightmoveID . "');";

        #Echo the query to check it

        echo $query_enter_rightmove_ID . "</br>";

        #Execute the query
        $query_enter_rightmove_ID = mysql_query($query_enter_rightmove_ID);
        echo "Leaving InsertRighmoveID() function </br>";

        #Execute the query
        $query_enter_rightmove_ID = mysql_query($query_enter_rightmove_ID);

        #Check to see if the query worked   
        if (!$query_enter_rightmove_ID) 
            {
                die("Database query failed:" . mysql_error());
            }
        echo "Leaving InsertRighmoveID() function </br>";
}

when I run the code in a webpage, get it to print the query to screen this is the message:

INSERT INTO tblRightMoveIDs (rightmoveID) VALUES ('44047607')

Database query failed:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

44047607 is the value passed to the function.

If I run the:

INSERT INTO tblRightMoveIDs (rightmoveID) VALUES ('44047607');

Outside the program it works.

Edit your code properly

Zword
– Zword

2013-11-21 10:12:14 +00:00
Commented Nov 21, 2013 at 10:12 — Zword
– Zword, Commented Nov 21, 2013 at 10:12
why u r making an easy program so complex?

R R
– R R

2013-11-21 10:20:00 +00:00
Commented Nov 21, 2013 at 10:20 — R R
– R R, Commented Nov 21, 2013 at 10:20

Shankar Narayana Damodaran · Accepted Answer · 2013-11-21 10:16:37Z

1

Remove the semicolon ; from here

 "INSERT INTO tblRightMoveIDs (rightmoveID)
    VALUES ('" . $RightmoveID . "');";
                            -------^

answered Nov 21, 2013 at 10:16

Shankar Narayana Damodaran

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Comments

Ramesh · Accepted Answer · 2013-11-21 10:18:41Z

0

There is error here in the starred parts:

**$query_enter_rightmove_ID** = mysql_query($query_enter_rightmove_ID);
echo "Leaving InsertRighmoveID() function </br>";
#Execute the query
$query_enter_rightmove_ID = mysql_query(**$query_enter_rightmove_ID**);

You are executing mysql_query first time with a proper query and it puts the result into $query_enter_rightmove_ID variable. Second time, you are using the result of first query as the parameter to mysql_query which is wrong

answered Nov 21, 2013 at 10:18

Ramesh

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