我仍在学习,任何人都可以帮助我,我的代码有什么问题?我需要加载,当您单击“加载”按钮时,程序将搜索在下拉列表中选择的数据库ID,它们将带名称..etc并显示在文本框中。对不起我的英语不好。
<?php
$servername = "localhost";
$username = "estgv15592";
$password = "estgv155922016";
$dbname = "estgv15592";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if(isset($_POST["loadbtn"]))
{
$id = (integer) $_POST["id"];
$query = "SELECT NOME, MORADA, PRECO FROM FICHA_DE_OBRA WHERE ID_FICHAOBRA = '$id' ";
$result = mysqli_query($conn, $query);
$details = mysql_fetch_array($result);
$nome = $details["NOME"];
$morada = $details["MORADA"];
$preco = $details["PRECO"];
}
$sql = "SELECT * FROM FICHA_DE_OBRA";
$result = mysqli_query($conn, $sql);
echo '<form id="form" method="post">';
echo "<select name ='id'>";
echo "<option value=''>Selecione Número ficha Obra</option>";
while($row = mysqli_fetch_array($result))
{
echo "<option value='" . $row['ID_FICHAOBRA'] . "'>" . $row['ID_FICHAOBRA'] . "</option>";
}
echo "</select>";
$conn->close();
?>
<input type="submit" value="Load" name="loadbtn">
<table width="300" border="0">
<tr>
<td>Name</td>
<td><input type="text" name="upName" style="text-align:right" value="<?php echo $nome;?>"/></td>
</tr>
<tr>
<td>Cost</td>
<td><input type="text" name="upCost" style="text-align:right" value="<?php echo $morada;?>" /></td>
</tr>
<tr>
<td>Active</td>
<td><input type="text" name="upActive" style="text-align:right" value="<?php echo $preco;?>" /></td>
</tr>
</table>
</div>
<br/>
</form>
您没有使用适当的php标签:(例如<?php echo $preco;?>
):
<tr>
<td>Name</td>
<td><input type="text" name="upName" style="text-align:right" value="<?php echo $nome; ?>"/></td>
</tr>
<tr>
<td>Cost</td>
<td><input type="text" name="upCost" style="text-align:right" value="<?php echo $morada; ?>" /></td>
</tr>
<tr>
<td>Active</td>
<td><input type="text" name="upActive" style="text-align:right" value="<?php echo $preco; ?>" /></td>
</tr>
使用mysqli_query
和mysqli_fetch_array
函数,请注意,其中的第一个参数mysqli_query
应该是您犯错的连接对象:
$result = mysqli_query($conn, $query); // first PHP block
$result = mysqli_query($conn, $sql); // second PHP block
$details = mysqli_fetch_array($result); // first PHP block
$row = mysqli_fetch_array($result) // second PHP block
然后$conn
将以下行移动到第一个PHP块的顶部,或者在第一个PHP块中未定义:
$servername = "localhost";
$username = "estgv15592";
$password = "your_password";
$dbname = "estgv15592";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
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