Question

我不明白为什么我不能插入数据库。我已经使用此表创建了一个数据库：

prenotazione（id，nome_rich，cognome_rich，email_rich，oggetto_rich）

interni（id，nome_int，cognome_int，email_int）

esterni（id，nome_est，cognome_est，email_est）

这是我的index.php

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 
Transitional//EN" 
"http://www.w3.org/TR/html4/loose.dtd">
<html>

<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<title>Prenotazione Videoconferenza</title>
<script src="utils.js"></script>
</head>

<body>



Inserire i dati richiesti:<br><br>
<form method="post" action="input.php">
<b> Richiedente Conferenza:</b><br><br>
Nome:<input type="text" name="name" size="20"><br>
Cognome:<input type="text" name="surname" size="20"><br>
Email: <input type="email" name="email" size="20"><br>
Oggetto Conferenza:<br><textarea name="testo" rows="5" cols="40" placeholder="Specificare oggetto Videoconferenza"></textarea><br>

<br>
<b>Partecipanti Interni</b>
<br>
<br>
<div id="start">
    <div id="first">
      Nome:<input type="text" name="iname[]" size="20"><br> 
      Cognome: <input type="text" name="isurname[]" size="20"><br> 
      Email: <input type="email" name="iemail[]" size="20"><br>
      <br>
    </div>
  </div>
  <br>
 Numero partecipanti interni:
 <input type="text" id="n1" value="1"><br>

 <button><a href="#" id="add">Aggiungi partecipante</a></button>



<br>
<b>Partecipanti Esterni</b>
<br>
<br>
Numero partecipanti Esterni:
 <input type="text" id="n2" value="1"><br>

 <button><a href="#" id="add2">Aggiungi partecipante</a></button>

  <div id="start2">
    <div id="first2">
      Nome:<input type="text" name="ename[]" size="20"><br> 
      Cognome: <input type="text" name="esurname[]" size="20"><br> 
      Email: <input type="email" name="eemail[]" size="20"><br>
      <br>
    </div>
  </div>
<input type="submit" value="Invia" > 
</form>
</body>
</html>

这是我用来插入数据（以及连接到DB）的input.php

<?php

$conn = @pg_connect("dbname=postgres user=postgres password=123456789");

if(!$conn) {
    die('Connessione fallita !<br />');
} else {
    echo 'Connessione riuscita !<br />';
}



// Richiedente
$name = $_POST['name'];
$surname = $_POST['surname'];
$email = $_POST['email'];
$testo = $_POST['testo'];

// Interni
if($_POST['iname']) foreach($_POST['iname'] as $iname) 
if($_POST['isurname']) foreach($_POST['isurname'] as $isurname)
if($_POST['iemail']) foreach($_POST['iemail'] as $iemail)

// Esterni
if($_POST['ename']) foreach($_POST['ename'] as $ename) 
if($_POST['esurname']) foreach($_POST['esurname'] as $esurname)    
if($_POST['eemail']) foreach($_POST['eemail'] as $eemail)


//inserting data order
INSERT INTO prenotazioni  (id,nome_rich, cognome_rich, email_rich,oggetto_rich)
            VALUES (1,'$name','$surname', '$email','$testo');


INSERT INTO interni (nome_int, cognome_int, email_int)
            VALUES ('$iname','$isurname','$iemail');

INSERT INTO esterni  (nome_est, cognome_est, email_est)
            VALUES  ('$ename','$esurname','$eemail');



?>

utils.js

$(document).ready(function() {
    $("#add").click(function(){

      var val1 =$("#n1").val();
      for(var i=0;i<val1;i++){
      $("#start").append($("#first").clone());
      }
    });
});

$(document).ready(function() {
    $("#add2").click(function(){

      var val2 =$("#n2").val();
      for(var i=0;i<val2;i++){
      $("#start2").append($("#first2").clone());
      }
    });
});

我已经测试了与数据库的连接，并且我收到了肯定的消息（在echo中）。但是如果我控制这些表，它们是空的。

Answer 1

尝试以下代码

// Richiedente
$name = $_POST['name'];
$surname = $_POST['surname'];
$email = $_POST['email'];
$testo = $_POST['testo'];
//Connecting to db here
$conn_string = "host=localhost port=5432 dbname=test user=lamb password=bar"; // change the db credentials here
$conn = pg_connect($conn_string);
//inserting data order
$query1 = "INSERT INTO prenotazioni  (id,nome_rich, cognome_rich, email_rich,oggetto_rich) VALUES (1,'$name','$surname', '$email','$testo')";
//execute the query here
$result = pg_query($conn, $query1 ); //if you are using pg_query and $conn is the connection resource
// Interni
$query = "";
if( !empty( $_POST['iname'] ) ) {

    foreach( $_POST['iname'] as $key => $iname ) {

        $isurname = empty( $_POST[$key]['isurname'] ) ? NULL : $_POST[$key]['isurname'];
        $iemail = empty( $_POST[$key]['iemail'] ) ? NULL : $_POST[$key]['iemail'];
        $query .= " ( '$iname', '$isurname', '$iemail' ) ";
    }
}
if( !empty( $query ) ) {

    $query2 = "INSERT INTO interni (nome_int, cognome_int, email_int) VALUES ".$query;
    $result = pg_query($conn, $query2 ); //if you are using pg_query and $conn is the connection resource
}
// Esterni
$query = "";
if( !empty( $_POST['ename'] ) ) {
    foreach( $_POST['ename'] as $key => $ename ) {
        $esurname = empty( $_POST[$key]['esurname'] ) ? NULL : $_POST[$key]['esurname'];
        $eemail = empty( $_POST[$key]['eemail'] ) ? NULL : $_POST[$key]['eemail'];
        $query .= " ( '$ename', '$esurname', '$eemail' ) ";
    }
}

if( !empty( $query ) ) {

    $query3 =  "INSERT INTO esterni  (nome_est, cognome_est, email_est) VALUES  " . $query;
    $result = pg_query($conn, $query3 ); //if you are using pg_query and $conn is the connection resource
}

此代码正试图将其批量插入数据库。尝试回显查询字符串，以查看创建的sql查询并在PG上运行该查询，以查看查询是否存在任何问题。

使用PHP和Postgresql插入数据库

1 个答案: