这个查询有什么问题

Question

if (!empty($_POST['comment']))
 {
       mysql_query("UPDATE contacts SET opp=1, SET inforecall='"$_POST['comment']"' WHERE contact_id='"$_GET['id']"' ");

}

Answer 1

Your first problem is lack of concatenation operators, change:

"UPDATE contacts SET opp=1, SET inforecall='"$_POST['comment']"' WHERE contact_id='"$_GET['id']"' ";

to:

"UPDATE contacts SET opp=1, inforecall='" . $_POST['comment'] . "' WHERE contact_id='" . $_GET['id'] . "' ";

Answer 2

replace your code with this

    if (!empty($_POST['comment']))

    {

    $id=$_GET['id'];

    $comment=$_POST['comment'];

    $query="UPDATE contacts SET opp=1, SET inforecall='$comment' WHERE contact_id='$id'";
               mysql_query($query);
}

Answer 3

Don't repeat SET . It's just

UPDATE tablename SET col1=val1, col2=val2 WHERE condition;

Answer 4

You should not give multiple SET keyword against each fieldname - The proper code should be:

if (!empty($_POST['comment'])) {
  mysql_query("UPDATE contacts SET opp=1, inforecall='"$_POST['comment']"' WHERE ontact_id='"$_GET['id']"' ");
}

Answer 5

if (!empty($_POST['comment']))
{
       mysql_query("UPDATE contacts SET opp=1, inforecall='{$_POST['comment']}' WHERE contact_id='{$_GET['id']}'");
}

That should solve your problems. If your contact_id is an INT then remove both of the single quotes after its equal sign.

Answer 6

首先，您不应该两次设置SET，而只需

UPDATE contacts SET opp=1, inforecall=...

Answer 7

First of all there are two set in the query, secondly there is an error in the syntax ( the php code $_post and $_get should be concatinated along with rest of the string and there are both get and post method used in the same place which are not sent at the same time.if you use post for both Maybe this will work,

if (!empty($_POST['comment']))
 {
       mysql_query("UPDATE contacts SET opp=1,inforecall='".$_POST['comment']."' WHERE contact_id='".$_POST['id']."' ");

}