我一直在努力做到这一点，我看了几本教程并阅读了数百页...我开始认为我想做的事是不可能的，也是不合逻辑的，即使我的脑海里也像是一个非常简单的概念
这是我的html / php代码
<?php
session_start();
include 'conect.php';
$consultar = "SELECT * FROM saida";
$resultado = mysqli_query($link,$consultar);
$sqlContrato = "SELECT * FROM saida WHERE NumeroContrato = (SELECT MAX(NumeroContrato) FROM saida);";
$resultado2 = mysqli_query($link,$sqlContrato);
?>
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="mota.css"/>
<title>Mota Saida</title>
<meta charset="utf-8">
</head>
<body id="main">
<header>
<h1 id="title">
Mota Saida
</h1>
</header>
<div id="container">
<p class="hidden">.</p>
<div id="pae">
<div id="form-div">
<form action="saida2.php" method="post" enctype="multipart/form-data">
<div class="child">
<div class="label-form">
<label for="numC">Numero de Contrato:</label>
</div>
<?php while($vector2=mysqli_fetch_array($resultado2)) { ?>
<div class="input">
<?php
echo $vector2[0]?>
</div>
</div>
<div class="child">
<div class="label-form">
<label for="nMota">Numero da mota</label>
</div>
<div class="input">
<?php
$NumerodaMota = $vector2[1];
$sqlMota = "SELECT * FROM bike WHERE NumeroMota = $NumerodaMota[0]";
$resultado3 = mysqli_query($link,$sqlMota);
echo $NumerodaMota;?>
</div>
</div>
<div class="child">
<div class="label-form">
<label for="dias">Dias</label>
</div>
<div class="input">
<?php echo $vector2[2]?>
</div>
</div>
<form action="saida2.php" method="post" enctype="multipart/form-data">
<div class="child">
<div class="label-form">
<label for="kilometrosS">Kilometros</label>
</div>
<div class="input">
<input type="number" name="numerohidden" hidden value='<?php
$vector2[0] ?>'>
在我的脑海中，这似乎行得通，但没有成功。
<input type="number" name="numerohidden" hidden value='<?php
$vector2[0] ?>'>
这是我的php / sql代码
<?php
define('SERVIDOR', 'localhost');
define('USUARIO', 'root');
define('BD', 'moto');
define('PASSWORD', '');
$link = mysqli_connect(SERVIDOR, USUARIO, PASSWORD, BD) or die("Error de conexion de base de datos");
if(!mysqli_select_db($link,'moto')){
echo 'db not selected';
echo mysqli_erro($link);
}
$NumeroContrato = $_POST['numerohidden'];
echo $NumeroContrato;
我正在尝试过去该数字，以便我可以更新sql表
$sql2 = "UPDATE saida SET Kilometros= '$Kilometros', NumeroHelmet1 = '$NumeroHelmet1',NumeroHelmet2 = '$NumeroHelmet2',DataSaida = '$DataSaida' ,HoraSaida = '$HoraSaida',DataEntrada = '$DataEntrada',HoraEntrada ='$HoraEntrada',Combustivel = '$Combustivel',Seguro = '$Seguro',ValorSeguro = '$ValorSeguro',Aluger = '$Aluger',Total = '$Total',Funcionaria = '$Funcionaria',Loja = '$Loja',Estado = '$Estado',Ticket = '$Ticket' WHERE NumeroContrato = '$NumeroContrato'" ;
我是php编程的新手，所以如果您可以帮助我，请这样做。对不起我的英语不好。
您的代码中有2个错误：1.输入类型不是“隐藏”的。2.您必须回显您的php变量。
更改
<input type="number" name="numerohidden" hidden value='<?php
$vector2[0] ?>'>
至
<input type="hidden" name="numerohidden" value='<?php echo $vector2[0] ?>'>
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