ricardogarcia Posted December 31, 2014 Share Posted December 31, 2014 (edited) Okay here is what I'm trying to do...--------------------------------------------------primary function of what I'm trying to do-----------------------------------------------------------simply....if the user picks a zipcode from $loc1 then image $image1(nextday) will show... if user pick a zipcode from $loc2 than $image2(2nd-Day)will hopefully appear under the search box..i hope that no page "redirect" is necessary... since images are displayed under search box area...---------------------------------------------------------------------PHP--Form-------------------------------------------------------------------------------<?php$zipcode = $_POST['zipcode'];$loc1 = array (80013,80015,80126,80127,80128,80129,80130,80134,80135,80138,80160,80161,80162,80163,80165,80166,804 33,80453,80465,80470);$loc2 = array (80023,80025,80136,80137,80138,80139,80140,80144,80145,80148,80140,80141,80142,80143,80145,80146,804 43,80443,80445,80440);// need to add whatever validation/sanitation of zip code value here $image1 = '<img src="Section-Images/Section-Edits/1st--Day-img.png" alt="1st--Day.png/>'; $image2 = '<img src="Section-Images/Section-Edits/2nd--Day-img.png" alt="2nd--Day.png" />'; $image3 = '<img src="Section-Images/Section-Edits/2-3--Day.png" alt="2-3--Day.png" />';if(in_array($zipcode, $loc1)) {echo $image1;}else if(in_array($zipcode, $loc2)) {echo $image2;}else {echo $image3;$form['#zipcode'] = FALSE;exit();}?>---------------------------------------------------------------------PHP--Form--End----------------------------------------------------------------------------------------------------------------------------------------------HTML--Form-------------------------------------------------------------------------------<div class="panel"><div class="panel-content"><form id="zipcode" action="<?php echo $_SERVER['PHP_SELF']; ?>" method='POST'><input type='text' size="6" maxlength="5" id='searchbar' class='searchbar2'name='zipcode'><div id="actions"><input type='submit' value='Search' onclick="submitclick(); return false;" id='submitbutton' class='zipcode_button'><td><id="zipcodeError" value='' class="zipcode_button"/></td></div></form></html>---------------------------------------------------------------------HTML--Form--End---------------------------------------------------------------------------- Edited December 31, 2014 by ricardogarcia Link to comment Share on other sites More sharing options...
Ingolme Posted December 31, 2014 Share Posted December 31, 2014 So where is the problem? I see you have a few variables that are not being used anywhere, namely $width, $height and $form. PHP scripts only run as the page loads, if you want to run them again you have to load the page again. This isn't valid HTML <td><id="zipcodeError" value='' class="zipcode_button"/></td> A <td> element shouldn't be here, and the other element <id="zipcodeError"> does not have a name. Link to comment Share on other sites More sharing options...
ricardogarcia Posted December 31, 2014 Author Share Posted December 31, 2014 This is really all i'm trying to do.. simply....if the user picks a zipcode from $loc1 then image $image1(nextday) will show... if user pick a zipcode from $loc2 than $image2(2nd-Day)will hopefully appear under the search box..i hope that no page "redirect" is necessary... since images are displayed under search box area... Link to comment Share on other sites More sharing options...
Ingolme Posted December 31, 2014 Share Posted December 31, 2014 If you don't want to redirect you'll have to use Javascript and AJAX. I understand what you want, but what part of your code is not working? Link to comment Share on other sites More sharing options...
ricardogarcia Posted December 31, 2014 Author Share Posted December 31, 2014 The part of the code that i'm having issues is that if; i load the page in browser, it displays some of the php code and image on top of template, (if(in_array($zipcode, $loc1)) { echo $image1; $width = 100; $height = 662; } else if(in_array($zipcode, $loc2)) { echo $image2; $width = 100; $height = 662; } else { echo $image3; $width = 100; $height = 662; $form['#zipcode'] = FALSE; exit(); } ?>) i'm not to sure if the code is correct but, ...i'm really just trying to shape the code just to do the following described, " if the user picks a zipcode from $loc1 then image $image1(nextday) will show... if user pick a zipcode from $loc2 than $image2(2nd-Day)will hopefully appear under the search box.. " thanks very much Link to comment Share on other sites More sharing options...
Ingolme Posted December 31, 2014 Share Posted December 31, 2014 It sounds like you're not running this on a PHP-enabled server or you forgot to save the file with a .php extension. Link to comment Share on other sites More sharing options...
proudly Posted February 27, 2015 Share Posted February 27, 2015 .. since images are displayed under search box area... I'm a newbie, but I think you , as Ingolme said , want to change only a part of your page. That is only achieved using AJAX Link to comment Share on other sites More sharing options...
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