2表单,首先搜索并在第二个表单中显示值

时间:2014-07-09 13:49:26

标签: php html

HTML页面:http://imgur.com/P95CyMv,VV4QYKS,xOE8RVA

id = 1的数据库行:http://imgur.com/P95CyMv,VV4QYKS,xOE8RVA#1

当我点击第一个表格的提交按钮时,我想在第二个表格的文本框中显示id = 1的值。

例如:http:// imgur.com / P95CyMv,VV4QYKS,xOE8RVA#2(不含空格)

我这样做,并且在工作前几个小时,但是我做了一些事情而现在还没有工作。

HTML / PHP:

<form method="post">
                <p>Mostrar evento por:</p>
                <p>
                    <select name="select" id="select">
                        <option selected="selected">id</option>
                        <option>Nome do Evento</option>
                        <option>Data do Evento</option>
                    </select>
                </p>
                <p>
                    <input type="text" name="nomeEvento" id="text" width="300">
                </p>
                <?php
                require ('sql_connect.php');

                if (isset($_POST['submit'])){
                    $selectedItem = mysql_escape_string($_POST['select']);
                    $nomeEvent = mysql_escape_string($_POST['nomeEvento']);

                    if (!$_POST['select'] | !$_POST['nomeEvento'])
                    {
                        echo ("<SCRIPT LANGUAGE='JavaScript'>
                            window.alert('Tem de meter um username e uma password')
                            window.location.href='UpdateEvents.php'
                            </SCRIPT>");
                        exit();
                    }

                    $query = "select * from `eventos` where `$selectedItem` = '$nomeEvent'";
                    $result = mysql_query($query);

                    if (!$result) die ("Database access failed: " . mysql_error());

                    while($row = mysql_fetch_array($result))
                    {
                        $coiso1 = $row['nome'];
                        $coiso2 = $row['data'];
                        $coiso3 = $row['descricao'];
                        $coiso4 = $row['url'];
                    }

                    echo $selectedItem, $nomeEvent, $coiso1, $coiso2, $coiso3;

                }                   
                ?>                
                <p>
                    <input type="button" name="Button" value="Mostrar" onClick="NomeEvento('<?php echo $coiso1 ?>','<?php echo $coiso2 ?>', '<?php echo $coiso3 ?>');">
                </p>
            </form>

            <form method="get">
              <p>Nome do Evento:</p>
              <p>
                <input name="text1" type="text" id="text1" width="300">
              </p>
              <p>Data do Evento:</p>
              <p>
                <input type="text" name="text" id="text2" width="300">
              </p>
              <p>Descrição:</p>
              <p>
                <input type="text" name="text2" id="text3" width="300">
              </p>
              <p>
                <input type="submit" name="submit2" value="Alterar">
              </p>
            </form></p>>

JAVASCRIPT FUNCTION:

<script>
        function NomeEvento(nomeEvento, dataEvento, descricaoEvento)
        {    
            document.getElementById("text1").value= nomeEvento;
            document.getElementById("text2").value= dataEvento;
            document.getElementById("text3").value= descricaoEvento;
        }

        </script>

有人可以帮助我吗?

1 个答案:

答案 0 :(得分:0)

目前,当您点击提交按钮时,您没有$coiso1$coiso2$coiso3个变量,因为您的php脚本只会在POST请求中填充它。

尝试在页面加载时调用方法NomeEvento(var1, var2, var3)