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I am trying to show the selected value in the list if matched it is successfully populated but not selected value code Have run 

  Code: $ employee name = 'Jimmy Chain'; & Lt; Select name = "employee" id = "employee" & gt; & Lt ;? Php $ data = array (); $ Data [0] = ''; Echo "& lt; option value = ''. $ Data [0]" '& gt; $. $ Data [0] "& lt; ($ No = $ DB- & gt; query) ('no first name, last name by staff'); foreign currency ($$$ data result) {$ SNo = $ data [' NO ']; $ SFN = $ data [' first name ']; $ SLN = $ data [' last name ']; $ SName = $ SFN.' '$ SLN. If ($ SName == $ employee name) {resonance "& Lt; Option value = '"$ sno."' Selected = \ "selected \" & gt; "$ Sname." & Lt; / Options & gt; \ N ";} Else {echo" & lt; Option value = '". $ SNo"' & gt; ". $ SName" & lt; / Option & gt; ";}}? & Gt; & lt; / select & gt;   
 The second is to run the statement but if the statement is not there I already put" selected "inside Please give advice.   
 
 
 Try it out:  
  foreach ( $ $ $ Result) {$ SNo = $ data ['no']; $ sfn = $ data ['first name']; $ SLN = $ data ['last name']; $ SName = trim ($ SFN) . '' .trim ($ SLN); if (stroller ($ SName) == strtolower ($ employee name)) {resonance "& lt; Option value = '"$ sno."' Selected = 'selected' & gt; "$ Sname." & Lt; / Option & gt; \ N ";} Else {echo" & lt; Option value = '". $ SNo"' & gt; ". $ SNAME" & lt; / Options & gt; Besides, you have $ Sname in the "if" statement instead of $ SName. I'm not sure if PHP will make a difference, but only Keep in mind. $ Sno and $ SNo are the same for variables.

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