php - Adding a second form from Ajax output -
i working on admin module db-project. user chooses table dropdown , jquery submits name of table php selects fields, composes , echoes it. works charm.
function ajax_search(){ $("#table_list").show(); var search_val=$("#tables").val(); $.post("./ajax/xxxxx.php", {searched_table : search_val}, function(data){ if (data.length>0){ $('#spinner_div').hide(); $("#table_list").html(data); } }) }
now add form under table adding 1 row. if hard code form on html-page , uses jquery like
function ajax_add(){ $("#table_list").show(); var add_val=$("#manufacturer").val(); $.post("./ajax/xxxxxx.php", $("#addform").serialize(), function(data){ if (data.length>0){ $('#spinner_div').hide(); $("#table_list").html(data); } }) }
i can catch in php , add row. create form after have submitted first form because there number of different tables different needs.
i have tried add form echo after building result table like
echo "<form method='post' name='addform' id='addform'> <input type='hidden' name='add'>....
the table shows in output , looks ok in chromes element inspector. form never submitted php because "submit" not caught by
$(document).ready(function(){ ..... $("#add_button").click(function(e){ $('#spinner_div').show(); e.preventdefault(); ajax_add(); }); });
is not possible add form after page has been read first time?
if have been unclear in describing problem best answer questions.
regards
mats
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