我有一个网页，其中包含一个输入他的名字 code_client
index.html
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8"/>
<meta name="viewport" content="width=device-width, initial-scale=1.0"/>
</head>
<body>
<form action="getLatLon.php" method="POST">
<fieldset>
code_client: <input id="code_client" name="code_client" type="text"> <br><br>
<br>
<input type="submit" value="Voir" name="Chercher">
</fieldset>
</form>
</body>
</html>
网页图片:
当我输入值时，我将其发送 (POST) 到 getLatLong.php 文件，结果显示客户端信息(code_client 的所有者)
getLatLon.php
<?php
$DB_Server = "localhost"; //MySQL Server
$DB_Username = "root"; //MySQL Username
$DB_Password = "123456"; //MySQL Password
$DB_DBName = "db_abc"; //MySQL Database Name
$DB_TBLName = "location"; //MySQL Table Name
$code_client=$_POST['code_client'];
$objConnect = @mysqli_connect($DB_Server, $DB_Username, $DB_Password);
$objDB = @mysqli_select_db($objConnect, $DB_DBName);
$strSQL = "SELECT * FROM `location` WHERE CodeClient='".$code_client."' ";
$objQuery = @mysqli_query($objConnect, $strSQL) or die("Couldn't execute query:<br>" . mysqli_error(). "<br>" . mysqli_errno());
$arrRows = array();
$arryItem = array();
while($arr = mysqli_fetch_array($objQuery)) {
$arryItem["Id"] = $arr["Id"];
$arryItem["Latitude"] = $arr["Latitude"];
$arryItem["Longitude"] = $arr["Longitude"];
$arryItem["CodeClient"] = $arr["CodeClient"];
$arrRows[] = $arryItem;
}
$DATA = $arrRows;
echo json_encode($arrRows);
?>
结果图片
我尝试将表单的结果发送到另一个 php 文件
testExportvr.php
<?php
include 'getLatLon.php';
echo json_encode($DATA);
?>
我尝试将表单的结果发送到另一个 php 文件，但直到现在我都做不到，它显示的结果与正确的结果不同
有谁知道我做错了什么？我需要表单将数据提交到文件，同时结果将显示在另一个文件中
最佳答案
你说你想将结果保存到文件中并在另一个页面中显示结果。 为此，您不需要名为“testExportvr.php”的第三个文件 你必须使用 php 文件处理(我假设你想将它保存为文本文件)
您只需要在“getLatLon.php”中进行更改 像这样
<?php
$DB_Server = "localhost"; //MySQL Server
$DB_Username = "root"; //MySQL Username
$DB_Password = "123456"; //MySQL Password
$DB_DBName = "db_abc"; //MySQL Database Name
$DB_TBLName = "location"; //MySQL Table Name
$code_client=$_POST['code_client'];
$objConnect = @mysqli_connect($DB_Server, $DB_Username, $DB_Password);
$objDB = @mysqli_select_db($objConnect, $DB_DBName);
$strSQL = "SELECT * FROM `location` WHERE CodeClient='".$code_client."' ";
$objQuery = @mysqli_query($objConnect, $strSQL) or die("Couldn't execute query:<br>" . mysqli_error(). "<br>" . mysqli_errno());
$arrRows = array();
$arryItem = array();
while($arr = mysqli_fetch_array($objQuery)) {
$arryItem["Id"] = $arr["Id"];
$arryItem["Latitude"] = $arr["Latitude"];
$arryItem["Longitude"] = $arr["Longitude"];
$arryItem["CodeClient"] = $arr["CodeClient"];
$arrRows[] = $arryItem;
}
$DATA = $arrRows;
$file = fopen("data.txt", "w") or die("Unable to open file!");
fwrite($file, json_encode($DATA));//in case if you want to save text as json format
fclose($file);
echo json_encode($arrRows);
?>
关于javascript - 将表单结果从 php 文件重定向到另一个文件，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/51575551/