在此應用程序中,我有一個下拉菜單,以便您可以選擇要查看的表格。使用Ajax我將變量$ tbl作爲一個整數進行推送,下面的代碼打印出與該整數對應的表。有些東西雖然不起作用,但我需要一雙額外的眼睛來幫助調試。使用下拉菜單顯示錶格和值
if($tbl == 9){$table = "person";}
$conn = new mysqli(DBHOST, DBUSER, DBPASS, DBNAME);
$query = mysqli_query($conn, "SELECT * FROM $table ") or die(mysqli_error($conn));
if($query){
echo '<table border="1"><thead><tr>';
$colnames = array();
while ($finfo = mysqli_fetch_field($query)) {
$name=$finfo->name;
array_push($colnames, $name);
echo "<th>".$name."</th>";
}
mysqli_free_result($query);
echo '</tr></thead><tbody>';
$count = 0;
echo $colnames[$count]; // this test prints the correct column name but it ends up being printed OUTSIDE the table for some reason
while ($row = mysqli_fetch_array($query, MYSQLI_BOTH)){
echo '<td>'.$row[$colnames[$count]].'</td>';
$count++;
}
echo "</tbody></table>";
}
最後while循環是假設呼應出每個字段的值,但<tbody>
留空它不執行。