Question

我有一个网页，其中包含一个输入，其名称为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>

网页网页的图像：

当我输入值时，我将其发送（POST）到getLatLong.php文件中，并因此显示客户端的信息（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);

?>

我尝试将模板的结果发送到另一个php文件，但是直到现在我还是做不到，它显示的结果与正确的结果不同

有人知道我在做什么错吗？我需要表格将数据提交到文件中，同时结果将显示在另一个文件中

Answer 1

您说的是要将结果保存到文件中并在另一页中显示结果。为此，您不需要名为“ testExportvr.php”的第三个文件。您必须使用php文件处理（我假设您要将其保存为文本文件）

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);

?>

将表单结果从php文件重定向到另一个

1 个答案: