我正在嘗試將php添加到表單中，以便從數據庫中填充它，以進行作業(yè)分配。我收到一個解析錯誤“解析錯誤：語法錯誤，/l7p1.php中第97行的文件意外結束”。盯著這個文件幾個小時后，我似乎找不到我的錯誤。在將它們全部更改為單引號之后，我本來在任何時候都得到了“意外（T_VARIABLE）”（希望有一些我不理解的地方，因為我不確定這是如何改變的）將錯誤移動到代碼的最后一行和“文件意外結尾”。這是php的第一部分，旨在用數據庫中的項目（工具）填充表單<?php $db=mysli_connect(null,null,null,'weblab') or die("Can't connect to db:".mysqli_connect_error($db)); $q = "select tool_item_no, tool_name, tool_price, tool_weight"; $q .= "from tool_t"; $q .= "order by tool_name"; $dbResult = mysqli_query($db,$q) or die("Database query error" . mysqli_error($db)); $num = mysqli_num_rows($dbResult); if ($num == 0){ echo"<h3>No Data</h3>"; } while ($row = mysqli_fetch_assoc($dbResult)) { $tool_item_no = $row["tool_item_no"]; $tmool_name = $row['tool_name']; $tool_price = $row['tool_price']; $tool_weight = $row['tool_weight']; echo "<label for = '$tool_name'>Name : $toolname<br/>Price : $tool_price<br/>Weight : $tool_weight<br/>How many $tool_name for this order? <input type = 'text' name = '$tool_name' id = '$tool_item_no' size = ]'5'/></label><br/><br/> }?>
3 回答
月關寶盒
TA貢獻1772條經驗 獲得超5個贊
以下應該工作。您混用了單引號，而您還不能設置Ca n't，所以需要將其括在“”中以轉義單引號。我還在您的php中添加了“”并添加了句點。您需要查找有關php變量和字符串的更多教程。
$db=mysli_connect(null,null,null,'weblab')
or die("Can't connect to db:".mysqli_connect_error($db));
$q = "select tool_item_no, tool_name, tool_price, tool_weight";
$q .= "from tool_t";
$q .= "order by tool_name";
$dbResult = mysqli_query($db,$q)
or die("Database query error" . mysqli_error($db));
$num = mysqli_num_rows($dbResult);
if ($num == 0){
echo"<h3>No Data</h3>";
}
while ($row = mysqli_fetch_assoc($dbResult)) {
$tool_item_no = $row["tool_item_no"];
$tmool_name = $row['tool_name'];
$tool_price = $row['tool_price'];
$tool_weight = $row['tool_weight'];
echo "<label for = ".$tool_name.">Name : ".$toolname".<br/>Price : ".$tool_price."<br/>Weight : ".$tool_weight."<br/>How many
".$tool_name." for this order?
<input type = 'text name = ".$tool_name." id = ".$tool_item_no." size = ]'5'/></label><br/><br/>"
}
$db=mysli_connect(null,null,null,'weblab')
or die("Can't connect to
db:".mysqli_connect_error($db));
$q = 'select state_abbr, state_name ';
$q .= 'from state_t ';
$q .= 'order by state_name;';
$dbResult = mysqli_query($db,$q)
or die('Database query error' .
mysqli_error($db));
$num = mysqli_num_rows($dbResult);
if ($num == 0){
echo '<option value=
'$state_abbr' >$state_name</option>';
echo '<option>Error</option>';
}
while ($row = mysqli_fetch_assoc($dbResult)) {
$state_abbr = $row[state_abbr];
$state_name = $row[state_name];
echo '<option value= '.$state_abbr.'>
'.$state_name.'</option>';
}
- 3 回答
- 0 關注
- 217 瀏覽
添加回答
舉報
0/150
提交
取消