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Update a Table

Posted on 2009-05-14
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Last Modified: 2012-05-07
I am learning php and need some help please.

I need to update record which i change, maybe the first record, maybe the last. I THINK i should have a loop, but KNOW i need help
if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "form1")) {

	$updateTable = array($_POST['Business_type'], $_POST['hidBusinessTypeID']);

		foreach($updateTable as $val) { 

  $updateSQL = sprintf("UPDATE business_type SET Business_type=%s WHERE Business_Type_id=%s",

                       GetSQLValueString($_POST['Business_type'], "text"),

                       GetSQLValueString($_POST['hidBusinessTypeID'], "int"));
 

  mysql_select_db($database_CallBack, $CallBack);

  $Result1 = mysql_query($updateSQL, $CallBack) or die(mysql_error());

}

}

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Question by:protype
  • 7
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12 Comments
 
LVL 39

Expert Comment

by:Roger Baklund
Comment Utility
If you are only updating one row, you don't need a loop.

How does the code you have work, do you get an error? Does it update the row?

Do you have this somewhere before the code you showed us:

$CallBack = mysql_connect('localhost','user','password');
$database_CallBack = 'name_of_database';

If so, this should work:
if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "form1")) {

  $updateSQL = sprintf("UPDATE business_type SET Business_type=%s WHERE Business_Type_id=%s",

                       GetSQLValueString($_POST['Business_type'], "text"),

                       GetSQLValueString($_POST['hidBusinessTypeID'], "int"));

  mysql_select_db($database_CallBack, $CallBack);

  mysql_query($updateSQL, $CallBack) or die(mysql_error());

}

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LVL 39

Expert Comment

by:Roger Baklund
Comment Utility
Note that I did not change anything in your code, just removed the unnecessary loop and the unnecessary assignment to $Result1.
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Author Comment

by:protype
Comment Utility
It is fine for the update of one record all the necessary code is in the file.

what i need to do is update records which have changes

i want to add a loop so that it will update a record which i change, the record could be the first, middle or last. could be 2 or 3 records, could be all records.

if there is a change update if no change dont update
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LVL 39

Expert Comment

by:Roger Baklund
Comment Utility
Is Business_Type_id the primary key in the business_type table? If so, there is only one row for each Business_Type_id, and only this one row will be updated.

If you need to update multiple rows at the same time, you must also have multiple input fields for Business_Type_id and Business_type. Do you have multiple fields in the HTML, two for each row? Can you show us the HTML?
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Author Comment

by:protype
Comment Utility
I display the fields via repeat.


                <?php do { ?>

                  <input name="hidBusinessTypeID" type="hidden" id="hidBusinessTypeID" value="<?php echo $row_rsBusinessType['Business_Type_id']; ?>" />

                  <input name="Business_type" type="text" class="TeleT" id="Business_type" value="<?php echo $row_rsBusinessType['Business_type']; ?>" />

                  <br />

                  <?php } while ($row_rsBusinessType = mysql_fetch_assoc($rsBusinessType)); ?></p>

            <p>&nbsp;                          </p>

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LVL 39

Expert Comment

by:Roger Baklund
Comment Utility
This gives you multiple fields with the same name and id, that won't work. Try the code below. Note that the hidden field is not needed anymore, because the id is stored within the square brackets of the name of the text field.
                <?php do { ?>

                  <input name="Business_type[<?php echo $row_rsBusinessType['Business_Type_id']; ?>]" type="text" class="TeleT" id="Business_type_<?php echo $row_rsBusinessType['Business_Type_id']; ?>" value="<?php echo $row_rsBusinessType['Business_type']; ?>" />

                  <br />

                  <?php } while ($row_rsBusinessType = mysql_fetch_assoc($rsBusinessType)); ?></p>

            <p>                           </p>

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LVL 39

Expert Comment

by:Roger Baklund
Comment Utility
To fetch the input and update the rows, do like this:
if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "form1")) {

  mysql_select_db($database_CallBack, $CallBack);

  foreach($_POST['Business_type'] as $key=>$value) {

    $updateSQL = sprintf("UPDATE business_type SET Business_type=%s WHERE Business_Type_id=%s",

                         GetSQLValueString($value, "text"),

                         GetSQLValueString($key, "int"));

    mysql_query($updateSQL, $CallBack) or die(mysql_error());

  }

}

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Author Comment

by:protype
Comment Utility
Ok this displays fine, thanks for the advice.

But the update is not working can you help with that please.
<?php do { ?>

                  <input name="Business_type[<?php echo $row_rsBusinessType['Business_Type_id']; ?>]" type="text" class="TeleT" id="Business_type_<?php echo $row_rsBusinessType['Business_Type_id']; ?>" value="<?php echo $row_rsBusinessType['Business_type']; ?>" />

                  <br />

                  <?php } while ($row_rsBusinessType = mysql_fetch_assoc($rsBusinessType)); ?></p>

            <p>                           </p>

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LVL 39

Accepted Solution

by:
Roger Baklund earned 500 total points
Comment Utility
Did you see my last post? Do you get errors?
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Author Comment

by:protype
Comment Utility
SORRY did not see the POST and Had to go out for a min.

All looks to work fine Thanks A MIL

Because i am learning I need advice about other things how can i stick with someone or is it possible
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Author Closing Comment

by:protype
Comment Utility
Thank you very much, it a pitty i don't have someone to help me learn this stuff. I'm to old to go back to school.
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LVL 39

Expert Comment

by:Roger Baklund
Comment Utility
I monitor the PHP zones, but after this question is closed, if you use the link that shows up "ask a related question" I will get an email notification.
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