loading value from database to html-select input

Question

I have this field saved in MYSQL database.

<select id="sel1" name="grpID">
  <option value="">select</option>
  <option value="G00">option1</option>
  <option value="G01">option2</option>
  <option value="G02">option3</option>
  <option value="G03">option4</option>
</select>


$grpID=$_POST['grpID'];
mysqli_query($db_Conx, "INSERT INTO users (grpID) VALUES ('$grpID')");

NOW, when I load this field, I want the saved value selected in select input box.

For example, if the saved value is "G02", I want the select input field loaded like this ![selection image][1

Is there a efficient way to do this instead of selecting it with if clauses for every and each case?

A switch statement.

worldofjr
– worldofjr

2014-12-28 12:33:43 +00:00
Commented Dec 28, 2014 at 12:33 — worldofjr
– worldofjr, Commented Dec 28, 2014 at 12:33

Adarsh Rajput · Accepted Answer · 2014-12-28 12:38:17Z

2

1) retrieve $grpID from database before <SELECT> tag.

2) Replace html-select with:

<select id="sel1" name="grpID">
   <option value="">select</option>
   <option value="G00" <?php if($grpID=="G00") echo "selected"; ?> >option1</option>
   <option value="G01" <?php if($grpID=="G01") echo "selected"; ?> >option2</option>
   <option value="G02" <?php if($grpID=="G02") echo "selected"; ?> >option3</option>
   <option value="G03" <?php if($grpID=="G03") echo "selected"; ?> >option4</option>
</select>

OR you can use loop also, if <option> tags are too many

answered Dec 28, 2014 at 12:38

Adarsh Rajput

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