Question

我有一个SELECT框，其中的值是从数据库填充的，下面是代码

<select name="ea_name" id="ea_name">
<option value="" selected="selected">Please Select...</option>
<?php 
require 'include/DB_Open.php';
$sql = "SELECT ea_name FROM ea_error ORDER BY ea_name";
$myData = mysql_query($sql) or die(mysql_error());
while ($row = mysql_fetch_array($myData)){
echo "<option value=\"$ea_name\">" . $row['ea_name'] . "</option>";
}
include 'include/DB_Close.php';
?>
</select>

现在我有一个jquery，这样当SELECT值改变时，它会从db中提取数据并将其填充到textarea

$("#ea_name").on("change", function() {
$.ajax({
url: "retrieve.php",
type: "POST",
data: {
ea_name: $(this).val()
},
success: function(data) {
$("#results").html(data);
}
});
});
});
});

这是我现在的查询，

<?php
require 'include/DB_Open.php';

$ea_name = $_POST['ea_name'];

$sql="SELECT * FROM ea_error WHERE ea_name = '" . $ea_name . "'";
echo $sql;

$myData = mysql_query($sql) or die(mysql_error());

//to count if there are any results
$numrow = mysql_num_rows($myData) ;

if($numrow == 0)
{
    echo "No results found.";
}
else
{
echo '<fieldset><legend><strong>Information</strong></legend>
<table width="619" border="0" align="center">
<tr><th scope="row">Error</th></tr>
<tr><th scope="row">Resolution</th></tr>
<tr><th scope="row">Contact/s</th></tr>';

while($info = mysql_fetch_array($myData)) 
{
echo "<form action='retrieve.php' method='post'>";
echo "<tr>";
echo  "<td align='center'>" . "<textarea readonly=readonly name=error cols=75 rows=10> " . $info['error'] . "</textarea></td>";
echo "</tr>";
echo "<tr>";
echo  "<td align='center'>" . "<textarea readonly=readonly name=resolution cols=75 rows=10> " . $info['resolution'] . "</textarea></td>"; 
echo "</tr>";
echo "<tr>";
echo  "<td align='center'>" . "<textarea readonly=readonly name=contacts cols=75 rows=10> " . $info['contacts'] . "</textarea></td>"; 
echo "</tr>"; 
echo "</form>";
}
}
echo "</fieldset>"; 

include 'include/DB_Close.php';
?>

我添加了echo $sql;所以我可以看到输出，这是我得到的： SELECT * FROM ea_error WHERE ea_name =''找不到结果。

我试图查看来源并即时看到：

注意：未定义的变量： C：\ xampp \ htdocs \ XXX \ view_transactions.php 中的ea_name 75
开关

第75行引用此echo "<option value=\"$ea_name\">" . $row['ea_name'] . "</option>";

我需要帮助理解和修复此错误...我刚接触php编码...谢谢

Answer 1

$ea_name未定义任何没有任何价值的地方我认为您正在寻找这个

echo "<option value=\"$row['ea_name']\">" . $row['ea_name'] . "</option>";

无法使用下拉值获取数据

1 个答案: