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PHP won't display dynamic table/form when called by Jquery

Posted: Thu Feb 16, 2012 7:25 pm
by pizzipie
Within a php program I have the following code which, if you run the program as a stand alone will produce a table/form.
If I call this module, revise.php, from an .html module with: $.post("revise.php", {id: revId, revData: revFlg, dName: dispName}, showResultsFromServer, "json"); the table/form will not be produced. You get a blank screen. Firebug shows the Response as:
generatedSource.png
with no errors. How can I get ths table/form to show??
revise.php run as a standalone you get this:
form_output.png
. The array is produced by php/mysql and is what will be encoded as JSON and sent to the call back function.


Thanks in advance for your help, RP


Code: Select all


foreach($rowsRead[0] as $key => $value) {
	if($value!="") $connectData[$key]=$value;
}

printf("<div id='who'>");

printf("<fieldset class='rev1'>");
printf("<legend>Revising Record %s </legend>", $revId);
printf("<form name='revContactForm' id='revContactForm' method='POST' action=''>\n");
printf("<table class='revForm' border='3' cellpadding='5'  bgcolor='#feede3' ><tr>\n");

printf("<td><input type='text' hidden='hidden' name='revFlg' id='revFlg' value='true'></input></td></tr>\n");
printf("<td><input type='text' hidden='hidden' name='revId' id='revId' value=%s></input></td></tr>\n",$revId);

foreach($rowsRead[0] as $key => $value) {

	if($key=='Name_ID' | $key=='Vendor_ID') {
		continue;
	};
		printf("<td><label for='%s'>%s</label></td>\n",strtolower($key), preg_replace("/_/"," ",$key));
		
		if($key=="Display_Name") {
		
    		printf('<td><input type="text" name="%s" onfocus="showDisplayName();" id="%s" value="%s"></input></td></tr>\n',
    		strtolower($key), strtolower($key), $value);		

			continue;
		};		
		
		
		if (contains($key, "Notes")) {
			printf("<td><textarea name='%s' id='%s'  rows=5 cols=81> '%s' </textarea></td></tr>\n",strtolower($key),
			 strtolower($key), $value);
			continue;
		};
	
		printf("<td><input type='text' name='%s' id='%s' value='%s'></input></td></tr>\n", strtolower($key), strtolower($key), $value);
		
	}; // for each	

printf("<tr><td><input type='submit' class='hov'  name='submit' value='Save Contact Data'></input></td></tr>\n");	
printf("</table></form></fieldset></div>\n");


printf("<div id='inputForm'></div>\n");  // <!-- inputForm -->\n");
     
 
 printf("<div id='controls'>\n");
 
	 //printf("<button type='button' class='hov' name='connect' onclick=window.location.href='ContactFrontEnd.php'>Go Back</button>\n");            
 	//printf("<button type='button' class='hov'  name='savContact' value='Save Contact Data' onclick=saveContact()>Save Contact</button>\n");
 	
  	//printf("<input type='submit' class='hov'  name='submit' id='submit' value='Save Contact Data'></input>\n");	
 printf("</div>\n");
 
 	
printf("<div id='rbox'>\n</div>\n");		
printf("<div id='rbox2'>\n</div>\n");


Re: PHP won't display dynamic table/form when called by Jque

Posted: Sat Feb 25, 2012 3:15 pm
by jraede
Can you show us "showResultsFromServer"? I'm assuming it's a function. Try logging the AJAX response to the console to make sure you're getting one - if so, then it's a problem with that showResultsFromServer function.

Re: PHP won't display dynamic table/form when called by Jque

Posted: Mon Mar 05, 2012 3:55 am
by lenin14
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