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如何从另一个PHP文件获取输入值？

[英]How can I get the input value, from another PHP file?

 原文  2016-07-23 10:25:04  7   4    php/ html/ mysql/ input/ google-visualization

问题描述

I want to update my SQL statement with an input text field and need to pass the value. 我想用输入文本字段更新SQL语句，并需要传递值。

I have an index.php file and a getDataC.php file. 我有一个index.php文件和一个getDataC.php文件。 I need to insert a value in the text field in index to update a SQL statement in getDataC to use a function in index with it. 我需要在索引的文本字段中插入一个值，以更新getDataC的SQL语句以在索引中使用函数。 How can I pass the value of the index field to getDataC to use it in index again? 如何将index字段的值传递给getDataC以再次在index中使用它？

Here my code, from index.php : 这是我的代码，来自index.php ： 

<body>
<h1 align="center">BigData - Gruppe 2</h1>
</br>
<table width="100%" align="center">
    <tr>
        <td width="25%" align="center">
            <h2>Bitte w&auml;hlen Sie eine Abfrage aus!</h2>
        </td>
        <td width="75%" align="center">
            <h2>Grafische Darstellung</h2>
        </td>
    </tr>
    <tr>
        <td width="40%">
            <hr>
            <h3 align="center">Abfragen:</h3> </br>
            <div>
                Einfluss von Bildung auf Fertilitätsrate (Japan)
                <button type="submit" id="actionA">Anzeigen</button>
            </div> </br>
            <div>
                Einfluss von Internet auf Fertilitätsraten (Japan)
                <button type="submit" id="actionB">Anzeigen</button>
            </div> </br>
            <div>
                Einfluss von Internet und Bildung auf die Anzahl der Frauen in
                Wirtschaft/Politik </br>
                <form > <label for="LandC">Wähle ein Land: <input type=text id="LandC" name="LandC"> </label></form>

                <button type="submit" id="actionC">Anzeigen</button>
            </div> </br>

And, here the code from getDataC.php : 并且，这是来自getDataC.php的代码： 

<?php
$con = mysqli_connect('localhost','root','','bigdata');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}

if( isset($_GET['submit']) )
{
//be sure to validate and clean your variables
$landC = htmlentities($_GET['LandC']);

}
if( isset($landC) ) {
 $k = $landC;
}
else $k='Germany';


$qry = "SELECT  * FROM facts WHERE facts.Country = '".  $k ."' AND Frauen_Fuehrungspositionen_Prozent IS NOT NULL";

$result = mysqli_query($con,$qry);

mysqli_close($con);

$table = array();
//Es´rstelt die Spaöten der Tabelle
$table['cols'] = array(
//Labels for the chart, these represent the column titles
array('id' => '', 'label' => 'Land', 'type' => 'string'),
    array('id' => '', 'label' => 'Jahr', 'type' => 'number'),

array('id' => '', 'label' => 'Staatsausgaben GDP per capita in Dollar', 'type' => 'number'),
array('id' => '', 'label' => 'Internetzugang in Prozent', 'type' => 'number'),

array('id' => '', 'label' => 'Frauen Fuehrungspositionen in Prozent', 'type' => 'number')

); 

$rows = array();
foreach($result as $row){
$temp = array();

//Values / Füllt die Spalten in der hier angegebenen Reihenfolge mit Zahlen
$temp[] = array('v' => (string) $row['Country']);
    $temp[] = array('v' => (int) $row['Year']); 

$temp[] = array('v' => (double) $row['Staatsausgaben_GDP_per_capita_in_Dollar']); 
    $temp[] = array('v' => (double) $row['Internetzugang_in_Prozent']); 

$temp[] = array('v' => (double) $row['Frauen_Fuehrungspositionen_Prozent']);
$rows[] = array('c' => $temp);

}   
$result->free();

$table['rows'] = $rows;

$jsonTable = json_encode($table, true);
echo $jsonTable;
?>

Now, the else statement of getDataC sets $k to 'Germany' since it seems that the text of the input field LandC doesn't get passed to getDataC . 现在， getDataC的else语句将$k设置$k 'Germany'因为似乎输入字段LandC的文本没有传递给getDataC 。

4 个解决方案

解决方案1
 0  2016-07-23 10:30:39

使用此标签更改您的表单标签<form action="HERE PUT THE PLACE OF TH FILE getDATAC.php" method="GET">在此<form action="HERE PUT THE PLACE OF TH FILE getDATAC.php" method="GET">

解决方案2
 0  2016-07-23 10:32:34

<form>

Should be: 应该： 

<form action="getDataC.php" method="GET">

Keeping the action the correct path of the file. 保持操作正确的文件路径。

解决方案3
 0  2016-07-23 11:56:49

Your HTML with Change (change your form action if your file getDataC.php is at another location ) : 带有Change的HTML（如果文件getDataC.php位于另一个位置，则更改表单操作）： 

<body>
<h1 align="center">BigData - Gruppe 2</h1>
</br>
<table width="100%" align="center">
    <tr>
        <td width="25%" align="center">
            <h2>Bitte w&auml;hlen Sie eine Abfrage aus!</h2>
        </td>
        <td width="75%" align="center">
            <h2>Grafische Darstellung</h2>
        </td>
    </tr>
    <tr>
        <td width="40%">
            <hr>
            <h3 align="center">Abfragen:</h3> </br>
            <div>
                Einfluss von Bildung auf Fertilitätsrate (Japan)
                <button type="submit" id="actionA">Anzeigen</button>
            </div> </br>
            <div>
                Einfluss von Internet auf Fertilitätsraten (Japan)
                <button type="submit" id="actionB">Anzeigen</button>
            </div> </br>
            <div>
                Einfluss von Internet und Bildung auf die Anzahl der Frauen in
                Wirtschaft/Politik </br>
                <form action="getDataC.php">
                <label for="LandC">Wähle ein Land: <input type=text id="LandC" name="LandC"> </label>
                <button type="submit" id="actionC">Anzeigen</button>
                </form>
            </div> </br>

Your getDataC.php file with change : 您的带有更改的getDataC.php文件： 

<?php
if(isset($_GET['LandC']) ){
//be sure to validate and clean your variables
$landC = htmlentities($_GET['LandC']);

}
if(isset($landC) ) {
 $k = $landC;
}
else $k='Germany';
$qry = "SELECT  * FROM facts WHERE facts.Country = '".  $k ."' AND Frauen_Fuehrungspositionen_Prozent IS NOT NULL";

$result = mysqli_query($con,$qry);

mysqli_close($con);

$table = array();
//Es´rstelt die Spaöten der Tabelle
$table['cols'] = array(
//Labels for the chart, these represent the column titles
array('id' => '', 'label' => 'Land', 'type' => 'string'),
    array('id' => '', 'label' => 'Jahr', 'type' => 'number'),

array('id' => '', 'label' => 'Staatsausgaben GDP per capita in Dollar', 'type' => 'number'),
array('id' => '', 'label' => 'Internetzugang in Prozent', 'type' => 'number'),

array('id' => '', 'label' => 'Frauen Fuehrungspositionen in Prozent', 'type' => 'number')

); 

$rows = array();
foreach($result as $row){
$temp = array();

//Values / Füllt die Spalten in der hier angegebenen Reihenfolge mit Zahlen
$temp[] = array('v' => (string) $row['Country']);
    $temp[] = array('v' => (int) $row['Year']); 

$temp[] = array('v' => (double) $row['Staatsausgaben_GDP_per_capita_in_Dollar']); 
    $temp[] = array('v' => (double) $row['Internetzugang_in_Prozent']); 

$temp[] = array('v' => (double) $row['Frauen_Fuehrungspositionen_Prozent']);
$rows[] = array('c' => $temp);

}   
$result->free();

$table['rows'] = $rows;

$jsonTable = json_encode($table, true);
echo $jsonTable;
?>

解决方案4
 0  2016-07-23 12:38:32

在要发送数据的位置创建一个php文件，并使用相同的名称。

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