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syntax error on line 53

syntax error on line 53.

I'm following a simple ajax>php>mysql example posted here http://openenergymonitor.org/emon/node/107

I've come across syntax error on line 53 and I don't know how to fix this?

<!---------------------------------------------------------------------------------------------
Example client script for JQUERY:AJAX -> PHP:MYSQL example
by Trystan Lea : openenergymonitor.org : GNU GPL

I recommend going to http://jquery.com/ for the great documentation there about all of this
---------------------------------------------------------------------------------------------->
<html>
  <head>
    <script language="javascript" type="text/javascript" src="jquery.js"></script>
  </head>
  <body>

  <!---------------------------------------------------------------------------------------------
  1) Create some html content that can be accessed by jquery
  ---------------------------------------------------------------------------------------------->
  <h2> Module No2 </h2>
  <h3>Output: </h3>
  <div id="output">this element will be accessed by jquery and this text will be replaced</div>

  <script id="source" language="javascript" type="text/javascript">

  $(function () 
  {

    //-------------------------------------------------------------------------------------------
    // 2) Send a http request with AJAX http://api.jquery.com/jQuery.ajax/
    //-------------------------------------------------------------------------------------------
    $.ajax({                                      
      url: 'api.php',                  //the script to call to get data          
      data: "",                        //you can insert url argumnets here to pass to api.php for example "id=5&parent=6"
      dataType: 'json',                //data format      
      success: function(rows)          //on recieve of reply // change from data to rows
      {
		for (var i in rows)
         {
          var row = rows[i];          

          var id = row[0];				//get id //var id = row[0];	
          var vname = row[1];			//get name // var vname = row[1];
  
	
        //--------------------------------------------------------------------------------------
        // 3) Update html content
        //--------------------------------------------------------------------------------------
       			// $('#output').html("<b>id: </b>"+id+"<b> name: </b>"+vname);     //Set output element html
        //recommend reading up on jquery selectors they are awesome http://api.jquery.com/category/selectors/
      $('#output').append("<b>id: </b>"+id+"<b> name: </b>"+vname)
                  .append("<hr />");
    } 
  } 
});

  </script>
   
  </body>
</html>  

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Redscrapbook
Asked:
Redscrapbook
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1 Solution
 
macarrillo1Commented:
There is a problem with your '(' and '{'
line 28 has ({
line 33 {
line 35 {
line 47 missing )
line 49 }
line 50 }

you are missing a total of  one }  and two ).
0
 
macarrillo1Commented:
It looks like you are missing the closing }); from the $.ajax

Sorry my mistake on line 47.  It is good as it is.
0
 
RedscrapbookAuthor Commented:
ok can you replace  a missing part? I can't see anything please?
Show it in codes of a missing part.
0
 
macarrillo1Commented:
Change line 52 to:

});

You need to close the function from lines 22 and 23 and close the $.ajax from line 28.
Both these statements need to be closed with });.  You only have one on line 51. Thus you need to add another on line 52.
0
 
RedscrapbookAuthor Commented:
That is a fastest response...in ten minutes.
0

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