[英]Modify jQuery and PHP code to pull through ID and text value into OPTION element
我有一個簡單的SELECT菜單:
<select class='form-control' id='builders' name='builder_id'>
<option value="1">Oracle</option>
<option value="2">SQL</option>
</select>
我想使用第一個SELECT的ID來更改第二個SELECT的ID為“ regions”的內容
這是我正在使用的jQuery代碼:
$(document).ready(function() {
$('#builders').change(function() {
var currentValue = $(this).val();
$.get("ajax_cats.php", {'builder_id': currentValue}, function(data) {
var regions = $.parseJSON(data).fld_label;
$('#regions').empty();
for (var i = 0; i < regions.length; i++) {
var regionOption = '<option value="'+regions[i]+'">';
regionOption += regions[i];
regionOption += '</option>';
$('#regions').append(regionOption);
}
});
});
});
這是ajax_cats.php的內容
$sql = "SELECT cats.fld_label
FROM tbl_b_cats cats
WHERE cats.fld_parent = ?
ORDER BY cats.fld_label";
/* initialise */
$stmt = $conn->stmt_init();
/* prepare */
if (!$stmt->prepare($sql)) {
throw new Exception("Error preparing statement: $stmt->error, SQL query: $sql");
}
/* bind */
if (!$stmt->bind_param('s', $_GET['builder_id'])) {
throw new Exception("Error binding parameter: $stmt->error");
}
/* execute & store */
$stmt->execute();
$stmt->store_result();
/* 404 for no results */
if ($stmt->num_rows == 0) {
header('HTTP/1.0 404 Not Found');
exit;
} else {
$regions = array();
/* get results */
$stmt->bind_result($fld_label);
while ($stmt->fetch()) {
$regions['fld_label'][] = $fld_label;
}
echo json_encode($regions);
}
/* free and close */
$stmt->free_result();
$stmt->close();
$conn -> close();
這樣很好。 我要嘗試解決的事情是,如何做,不能解決,以便代替第二個SELECT在option元素的“ value”和display部分中包含相同值的東西:
<select class="form-control" id="regions" name="regions">
<option value="BLOB_NAV">BLOB_NAV</option>
<option value="Insert">Insert</option>
<option value="Select">Select</option>
</select>
我該如何更改代碼,以便可以進入例如
<select class="form-control" id="regions" name="regions">
<option value="1">BLOB_NAV</option>
<option value="2">Insert</option>
<option value="4">Select</option>
</select>
這樣,當基於“區域”選擇的值進行處理時,我可以使用內部ID代替文本值。
我可以看到我需要將ajax文件上的PHP更改為:
$sql = "SELECT cats.fld_label
, cats.fld_id
FROM tbl_b_cats cats
WHERE cats.fld_parent = ?
ORDER BY cats.fld_label";
/* initialise */
$stmt = $conn->stmt_init();
/* prepare */
if (!$stmt->prepare($sql)) {
throw new Exception("Error preparing statement: $stmt->error, SQL query: $sql");
}
/* bind */
if (!$stmt->bind_param('s', $_GET['builder_id'])) {
throw new Exception("Error binding parameter: $stmt->error");
}
/* execute & store */
$stmt->execute();
$stmt->store_result();
/* 404 for no results */
if ($stmt->num_rows == 0) {
header('HTTP/1.0 404 Not Found');
exit;
} else {
$regions = array();
/* get results */
$stmt->bind_result($fld_label, $fld_id);
...
echo json_encode($regions);
}
/* free and close */
$stmt->free_result();
$stmt->close();
$conn -> close();
但是后來我被困在:
<option>
元素中的正確位置 PHP:將每個記錄的關聯數組添加到結果數組。
while ($stmt->fetch()) {
$regions []= [
'id' => $fld_id,
'label' => $fld_label
];
}
JS:訪問每個JSON記錄的字段。
for (var i = 0; i < regions.length; i++) {
var regionOption = '<option value="'+regions[i]['id']+'">';
regionOption += regions[i]['label'];
regionOption += '</option>';
$('#regions').append(regionOption);
}
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