PHP drop down coding question -

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Can someone show me how to make a statement where the default value of 0 below a drop (my case ) Automatically submit form if possible? 

  session_start (); $ Current = release ($ _ session ['client namefor'])? $ _SESSION ['client namefor']: 0; While ($ line = mysql_fetch_array ($ result)) {$ id = $ row ["Client_Code"]; $ Cheese = $ line ["Client_Full_Name"]; $ Value = "$ id, $ thing"; ? $ Sel = ($ id == $ current) selected ':' '; $ Options4 = "& Lt; OPTION $ sel VALUE = \" $ value \ "& gt; $ cheese & lt; / option & gt;"; }? & Gt; & Lt; Form name = "form" action = "& lt ;? php echo $ _SERVER ['PHP_SELF'] ;? & gt;" Method = "post" & gt; & Lt; SELECT NAME = "ClientNameFour" OnChange = "this.form.submit ()") & gt; Option Price = 0> Customer & lt ;? Php $ options4? & Gt; & lt; / SELECT & gt; & lt; / FORM>;    
 
 
  onchange  event and some javascript Generally, your generated HTML should look something like this: 
 
  form id = "form" name = "form" method = "POST" action = "http : //example/script.php "& gt; select id =" select "name =" select "onChange =" document.getElementById ('form'). Submit (); "& Gt; Option value =" 0 "selected =" selected "> 0  <
 
 Compared to your solution: 
 
  onchange  event is only removed when you select an unselected option, no additional check is needed. 
 
 
 
     
  •   id  is missing in your  
     gt;  ta  
  •  you    Need to close the tag 
    •  
  •  you probably go Askript change in the  onchange   & lt; select & gt;  tag 
    •  
 
  Specialty

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