MySql + php + checkboxes = agghhh

Anon

Hi,

Im using a MySql database and php front end.

My problem :

I have a form that adds data into the database.

But i now want to recall the data from the database and display it on another form.

I have checkboxes on the form which send yes for ticked and no for unticked to the database.

Is there any way to check the value of the checkbox from the database ( yes or no) and then tick the tick box on the form if \"yes\" or leave it unchecked if \"no\" ???

Thanks for n e help in advance
Simon

Anon

Hello, how to export mysql DB to a file that can use in other sql application ?? I'm new in mysql, please tell me the step.

Thank You

Anon

Here is some sample code that should anwser your question. Assume that you have executed a query, and stored the value of the field "is_php_fan" in
the host variable $is_php_fan.

To check the corresponding HTML checkbox, do the following.

Regards,

Ulrik

Anon

Put CHECKED as an attribute in your input tag, dependent on an if statement.
eg:
if ($databaseitem == "yes") {
echo ("<input type=\"checkbox\" name=\"name\" value=\"value\" checked>");
}
else {
echo ("<input type=\"checkbox\" name=\"name\" value=\"value\">");
}

Where $databaseitem has been set to the relevant contents in your database.

I'm new to this PHP thing, so no promises.
Hope this helps.

Anon

Two lines of code...

// 0 = no and 1 = yes
if ($option == 0) {$checked = "";} else {$checked="checked"};
printf("<input type=\"checkbox\" name=\"name\" value=\"value\" %s>", $checked);

Anon

// 0 = no and 1 = yes
if ($option == 0) {$checked = "";} else {$checked="checked"};
printf("<input type=\"checkbox\" name=\"name\" value=\"value\" %s>", $checked);

Dropdown code below...

<?php
$values[0] = "Newbie";
$values[1] = "Regular";
$values[2] = "Veteran";

// Put the call below in your form
xDropDown("status",$values,1); 

// Put this function in your icommon.php file...

function xDropDown($name, $values, $default) {
// This function is called to display a drop down box
// $values is an array thats loaded prior to the calling of this function and
// contains a list of text will appear in the dropdown

printf("<select size=\"1\" name="%s">", $name);
for ($x=0;$x<count($values);$x++) { 
	if ($x == $default) {$selected = "checked";} else {$selected = "";}
	printf("<option %s value=\"%s\">%s</option>",$selected, $x, $value[$x];
}
printf("</select>");

}

Then, after the user submits the form, $status will contain either 0, 1 or 2 and you insert this into the MySql db column.

Initially, prior to calling xDropDown, you can query the database and load the previously selected value into the $default var... This causes the drop down to initially display the chosen db value... I've got more...

Anon

sorry i dont know some one else setup my Mysql database for me