按鈕單擊時,我正在使用ajax請求POSTSELECT數據到我的PHP頁面。出於某種原因,我無法讓我的PHP IF語句評估爲真。它的默認條件爲business_unit_brand = 3。我嘗試echoing和print_r $ brand_bu變量,看看它沒有運氣。
形式:
<select id="brand_bu" name="selected" class="form-control">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select>
<span class="input-group-btn">
<button class="btn btn-success" id="submitbu" type="button" tabindex="-1" action=""><span class="glyphicon glyphicon-retweet" aria-hidden="true"></span></button>
</span>
jQuery的阿賈克斯
$("#submitbu").click(function(event) {
console.log("Chose BU: " + $("#brand_bu").val());
$.ajax({
url: "table.php",
type: "POST",
dataType:'json',
data: JSON.stringify({'bu': $('#brand_bu').val()}),
success: function(data){ console.dir(data); refreshTable(); },
error: function(){alert("Something went wrong, please close the page and re-open.")}
}); });
PHP:
$brand_bu = $_POST['bu'];
if ($brand_bu == "1"){
$business_unit_brand = "1";
} else if ($brand_bu == "2"){
$business_unit_brand = "2";
} else if ($brand_bu == "3"){
$business_unit_brand = "3";
} else if ($brand_bu == "4"){
$business_unit_brand = "4";
} else if ($brand_bu == "5"){
$business_unit_brand = "5";
} else {
$business_unit_brand = "3";
}
這工作。感謝您的及時響應! – khush123456