[英]Loop inside PHP script are not executed
我有一個要在表單輸入標簽中顯示invoice_id的表單。 發票ID是從php腳本計算得出的,然后將其進一步通過變量傳遞給Java腳本,以獲取等於傳遞的變量的輸入值。 但是我不知道我在哪里犯錯,我正在獲取在其他條件下在php代碼中定義的值,不知道為什么php代碼循環不會執行。 我的桌子也已經有2行了。 php代碼是:
<?php
$conn = new mysqli("localhost","root","","tssolutions");
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sqlinvoice = "SELECT `invoiceid` FROM `transactiona` ORDER by `invoiceid` DESC LIMIT 1";
$resultinvoice = $conn->query($sqlinvoice);
echo "<br><br><br><br><br><br><br><br><br><br><br><br>";
if ($resultinvoice->num_rows != 0) {
// output data of each row
while($row = $resultinvoice->fetch_assoc()) {
echo $row['invoiceid'];
$row['invoiceid'] +=1;
echo $row['invoiceid'];
echo " <script> var invoid = ".$row['invoiceid']." </script>";
}
}
else {
echo " <script> var invoid = '1'; </script>";
}
$conn->close();
?>
HTML代碼是:
<form action="./manipulate/invoice.php" method="post" accept-charset="UTF-8">
<div style="width:15%;float:left">
Invoice #
<br>
<input style="width:98%;margin-left:1%;margin:right:1%" type="text" placeholder="Invoice ID" id="lmn">
</div>
</form>
<script>
$(document).ready(function () {
$("#lmn").val(invoid);
});
</script>
mysql_fetch_object()返回一個對象,而不是對象數組。 所以我要做的就是添加$ row = array(); 在此,感謝所有幫助我達成解決方案的人。
if ($resultinvoice->num_rows > 0) {
// output data of each row
$row = array();
while($row = $resultinvoice->fetch_assoc()) {
$row['invoiceid'] +=1;
echo " <script> var invoid = ".$row['invoiceid']." </script>";
}
}
else {
echo " <script> var invoid = '1'; </script>";
}
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