单击“添加”按钮时会弹出一个对话框。只要在对话框中按下提交按钮,它就会将其插入数据库。有两个字段MR_ID
和Supp_ID
。 Supp_ID
仅接受整数。我还要向Supp_ID
添加更多验证,以便在输入的值为重复值并且已在数据库中时不会提交或插入
我认为最好的方法是使用AJAX。单击提交按钮后如何向服务器写入AJAX调用并检查数据库中是否存在具有相同值的记录?
JavaScript的:
$( function() {
$("#insertButton").on('click', function(e){
e.preventDefault();
});
var dialog, form,
mr_id_dialog = $( "#mr_id_dialog" ),
supplier_id = $( "#supplier_id" ),
allFields = $( [] ).add( mr_id_dialog ).add( supplier_id ),
tips = $( ".validateTips" );
console.log(allFields);
function updateTips( t ) {
tips
.text( t )
.addClass( "ui-state-highlight" );
setTimeout(function() {
tips.removeClass( "ui-state-highlight", 1500 );
}, 500 );
}
function checkRegexp( o, regexp, n ) {
if ( !( regexp.test( o.val() ) ) ) {
o.addClass( "ui-state-error" );
updateTips( n );
return false;
} else {
return true;
}
}
function addVendor() {
var valid = true;
allFields.removeClass( "ui-state-error" );
// ----- Validation for each input in add row dialog box -----
valid = valid && checkRegexp( supplier_id, /^(0|[1-9][0-9]*)$/, "Please enter a valid Supplier ID" );
console.log(allFields);
if ( valid ) {
var $tr = $( "#index_table tbody tr" ).eq(0).clone();
var dict = {};
var errors = "";
$.each(allFields, function(){
$tr.find('.' + $(this).attr('id')).html( $(this).val()+"-"+supplier_id );
var type = $(this).attr('id');
var value = $(this).val();
console.log(type + " : " + value);
// ----- Switch statement that provides validation for each table cell -----
switch (type) {
case "mr_id_dialog":
dict["MR_ID"] = parseInt(value);
console.log(dict['MR_ID']);
break;
case "supplier_id":
dict["Supp_ID"] = value;
break;
}
});
$( "#index_table tbody" ).append($tr);
dialog.dialog( "close" );
console.log(dict);
var request = $.ajax({
type: "POST",
url: "insert.php",
data: dict
});
request.done(function (response, textStatus, jqXHR){
if(JSON.parse(response) == true){
console.log("row inserted");
} else {
console.log("row failed to insert");
console.log(response);
}
});
// Callback handler that will be called on failure
request.fail(function (jqXHR, textStatus, errorThrown){
console.error(
"The following error occurred: "+
textStatus, errorThrown
);
});
// Callback handler that will be called regardless
// if the request failed or succeeded
request.always(function () {
});
}
return valid;
}
var dialog = $( "#dialog-form" ).dialog({
autoOpen: false,
height: 400,
width: 350,
modal: true,
buttons: {
"Add Supplier ID": addVendor,
Cancel: function() {
dialog.dialog( "close" );
}
},
close: function() {
form[ 0 ].reset();
allFields.removeClass( "ui-state-error" );
}
});
form = dialog.find( "form" ).on( "submit", function( event ) {
event.preventDefault();
addVendor();
});
$( "#insertButton" ).button().on( "click", function() {
dialog.dialog({
position: ['center', 'top'],
show: 'blind',
hide: 'blind'
});
dialog.dialog("open");
});
});
我的对话框代码:
<div id="dialog-form" title="Add Supplier ID">
<p class="validateTips">All form fields are required.</p>
<!-- Dialog box displayed after add row button is clicked -->
<form>
<fieldset>
<label for="mr_id">MR_ID</label>
<select name="mr_id" id="mr_id_dialog" class="text ui-widget-content ui-corner-all" value="300">
<?php foreach($user1->fetchAll() as $user2) { ?>
<option>
<?php echo $user2['MR_ID'];?>
</option>
<?php } ?>
</select><br><br>
<label for="supplier_id">Supplier ID</label>
<input type="text" name="supp_id" id="supplier_id" class="text ui-widget-content ui-corner-all" value="99">
<!-- Allow form submission with keyboard without duplicating the dialog button -->
<input type="submit" id="submit" tabindex="-1" style="position:absolute; top:-1000px">
</fieldset>
</form>
</div>