我有兩個下拉列表,這些值是從MySQL填充的。第二個下拉值取決於第一個下拉選項。如何使用JavaScript在下拉菜單中顯示選項名稱
無論如何,代碼正在工作。現在使用我的代碼,我將hospital_id發佈到另一個php。但是我想在下拉菜單中顯示hospital_name作爲文本,但截至目前,我只能顯示hospital_id。
下面請看看我的代碼,並建議我一個解決方案:
$query = "SELECT bp_id,bp_name FROM mfb_billing";
$result = $db->query($query);
while($row = $result->fetch_assoc()){
$categories[] = array("bp_id" => $row['bp_id'], "val" => $row['bp_name']);
}
$query = "SELECT bp_id, hospital_id, hospital_name FROM mfb_hospital";
$result = $db->query($query);
while($row = $result->fetch_assoc()){
$subcats[$row['bp_id']][] = array("bp_id" => $row['bp_id'], "val" => $row['hospital_id']);
}
$jsonCats = json_encode($categories);
$jsonSubCats = json_encode($subcats);
這是腳本:
<script type='text/javascript'>
<?php
echo "var categories = $jsonCats; \n";
echo "var subcats = $jsonSubCats; \n";
?>
function loadCategories(){
var select = document.getElementById("categoriesSelect");
select.onchange = updateSubCats;
for(var i = 1; i < categories.length; i++){
select.options[i] = new Option(categories[i].val,categories[i].bp_id);
}
}
function updateSubCats(){
var catSelect = this;
var catid = this.value;
var subcatSelect = document.getElementById("subcatsSelect");
subcatSelect.options.length = 0; //delete all options if any present
for(var i = 0; i < subcats[catid].length; i++){
subcatSelect.options[i] = new Option(subcats[catid][i].val,subcats[catid][i].hosp);
}
}
</script>
這是我的形式:
<body onload='loadCategories()'>
<form id="reportvalue" action="testpj2.php" method="post">
<select id='categoriesSelect'>
<option value="1">Select Billing Provider</option>
</select>
<select name='hospitalname[]' id='subcatsSelect' multiple='multiple'>
<option value="all">Select Billing Provider</option>
</select>
<label for="from">From</label>
<input type="text" id="from" name="from">
<label for="to">to</label>
<input type="text" id="to" name="to">
<?php
//$a = $_REQUEST['hospitalname[]'];
//echo $a;
?>
<input type="submit" name="Submit" value="Submit">
</form>
你可以用兩種或兩種以上的方式來完成,將下拉列表的名稱設置爲一個隱藏的輸入,然後在POST中輸入值,或者通過在php –