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我的編碼是所有關於ALERT,當搜索詞不是在MySQL數據庫中找到
1)從MySQL獲取數據救援人員到場PHP
2)從PHP獲取數據使用PHP URL D3基於對輸入 我想設置的警報時,在輸入字段中的文本在MySQL數據庫中找不到..
現在當我嘗試用比MySQL數據之外的字,它顯示
此控制檯
我怎麼可以提醒時,錯一個字(比MySQL數據庫值等)被提交
HTML表單
<form name="editorForm">
<input type="text"name="editor" id="editor"
onchange="document.getElementById('editorForm').submit();">
<input type="submit"value="butn">
</form>
jQuery來來去PHP根據數據URL
$(function() {
$('form').submit(function (e) {
e.preventDefault();
var t=$('form').serialize();
var u='http://localhost:8888/saff/indexi.php?'+t;
if(u==null){
alert("not found");
}
else{
funn();
}
個
D3碼
function funn(){
d3.json(u, function(treeData) {
//D3 CODES
});
}
我的PHP代碼
<?php
$con=mysqli_connect("localhost","root","admin","data");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$name=$_GET['editor'];
$sql="SELECT * FROM phptab where value LIKE '%".$name."%'";
$r = mysqli_query($con,$sql);
$data = array();
while($row = mysqli_fetch_assoc($r)) {
$data[] = $row;
}
function buildtree($src_arr, $parent_id = 0, $tree = array())
{
foreach($src_arr as $idx => $row)
{
if($row['parent'] == $parent_id)
{
foreach($row as $k => $v)
$tree[$row['id']][$k] = $v;
unset($src_arr[$idx]);
$tree[$row['id']]['children'] = buildtree($src_arr, $row['id']);
}
}
ksort($tree);
return $tree;
}
function insertIntoNestedArray(&$array, $searchItem){
if($searchItem['parent'] == 0){
array_push($array, $searchItem);
return;
}
if(empty($array)){ return; }
array_walk($array, function(&$item, $key, $searchItem){
if($item['id'] == $searchItem['parent']){
array_push($item['children'], $searchItem);
return;
}
insertIntoNestedArray($item['children'], $searchItem);
}, $searchItem);
}
$nestedArray = array();
foreach($data as $itemData){
//$nestedArrayItem['value'] = $itemData['value'];
$nestedArrayItem['id'] = $itemData['id'];
$nestedArrayItem['name'] = $itemData['name'];
$nestedArrayItem['parent'] = $itemData['parent'];
$nestedArrayItem['tooltip'] = $itemData['tooltip'];
$nestedArrayItem['color'] = $itemData['color'];
$nestedArrayItem['level'] = $itemData['level'];
$nestedArrayItem['children'] = array();
//$data[]=$dat;
insertIntoNestedArray($nestedArray, $nestedArrayItem);
}
header('Content-Type: application/json');
$json= json_encode($nestedArray,JSON_UNESCAPED_UNICODE);
echo $json = substr($json, 1, -1);
?>
按預期工作時所使用的詞是在數據庫 並在Mozilla的page looks like this 得到正確的JSON格式存在console.but設計不顯示在頁面中...但在鉻,一切工作正常..
等待什麼呢?從網址'var u ='http:// localhost:8888/saff/indexi.php?'+ t;'獲取數據?你聽說過阿賈克斯嗎? – madalinivascu
yes.why使用這個url獲取數據時出現任何錯誤 – prog
當我在數據庫中的輸入字段中鍵入單詞時..正確獲取數據,沒有問題...當我鍵入不在它顯示的數據庫在控制檯JSON輸入的意外結束...但我想提醒「未找到」.. – prog