嘗試將下拉列表中的數據作為外鍵插入另一個表中。 數據未插入
[英]Trying to insert data from a drop-down list as a foreign key in another table. Data not inserted
問題描述
我一直試圖找出我的錯誤在哪里沒有成功。 我沒有報告任何錯誤。 我想將數據從PHP表單插入供應商表。 id_drugstore是supplier表中的外鍵，可以從下拉列表中進行選擇（ 下拉列表很完美 ）。 除id_supplier之外，其他日期由用戶id_supplier 。
<?
error_reporting(-1);
$conn = mysqli_connect("localhost", "root", "", "retete");
if(mysqli_connect_errno())
{
echo "Nu ma pot conecta la baza de date medical:" .mysqli_connect_error();
}
if(isset($_POST['submit']){
$id_drugstore=$_POST['id_drugstore'];
$name_suplier=$_POST['name_suplier'];
$country_supplier=$_POST['county_supplier'];
$county_supplier=$_POST['county_supplier'];
$town_supplier=$_POST['town_supplier'];
$street_supplier=$_POST['street_supplier'];
$bank_suplier=$_POST['bank_suplier'];
$no_cont_Bank=$_POST['$no_cont_Bank'];
$qry=mysqli_query("INSERT INTO supplier `VALUES('sss','%$id_drugstore%',$_POST['name_suplier']','$_POST['county_supplier']','$_POST['county_supplier']','$_POST['town_supplier']','$_POST['street_supplier']','$_POST['bank_suplier']','$_POST['$no_cont_Bank']";`
$inserted=($qry,$conn);
if($inserted){
echo "Datele au fost inserate cu suucess";
}else{
echo "datele nu au putut fi salvate in baza de date".mysqli_error($conn);
}
}
mysqli_close($conn);
}
?>
<body>
<div class="jumbotron">
<div class="container">
<h4 class="display-5">Adaugare Furnizori in baza de date</h4>
<p class="lead">Completati formularul de mai jos</p>
</div>
</div>
<form action="AdaugareFurnizori.php" method="POST">
<div class="form-row">
<div class="col-md-2 mb-3">
<label for="validationServer01">Farmacie</label>
<select name="name_drugstore">
<option>Selecteaza farmacie</option>
<?php
$conn = mysqli_connect("localhost", "root", "", "retete");
if(mysqli_connect_errno()){
echo "Nu ma pot conecta la baza de date medical:".mysqli_connect_error();
}
$q=mysqli_query($conn,"SELECT * FROM drugstore")or die(mysqli_error());
$c=mysqli_num_rows($q);
if($c==0){
echo 'Nu exista farmacie in baza de date';
}else{
while($row=mysqli_fetch_array($q)){
$id_drugstore=$row['id_drugstore'];
$name_drugstore=$row['name_drugstore'];
echo "<option value='$id_drugstore'>$id_drugstore/$name_drugstore<option>
}
}
mysqli_close($conn);
?>
</select>
</div>
</div>
<div class="form-row">
<div class="col-md-4 mb-3">
<label for="validationServer01">Denumire Furnizori</label>
<input type="text" name="name_suplier" class="form-control is-valid" id = "validationServer01" placeholder="Denumire Furnizor" required>
</div>
</div>
<div class="form-row">
<div class="col-md-2 mb-3">
<label for="validationServer02">Tara</label>
<input type="text" name="country_supplier" class="form-control is-valid" id="validationServer02" placeholder="Tara" required>
<div class="invalid-feedback">
Va rugam sa introduceti tara.
</div>
</div>
<div class="col-md-2 mb-3">
<label for="validationServer03">Judet</label>
<input type="text" name="county_supplier" class="form-control is-valid" id="validationServer03" placeholder="Judet" required>
<div class="invalid-feedback">
Va rugam sa introduceti judetul.
</div>
</div>
<div class="col-md-2 mb-3">
<label for="validationServer04">Oras</label>
<input type="text" name="town_supplier"class="form-control is-valid" id="validationServer04" placeholder="Oras" required>
<div class="invalid-feedback">
Va rugam sa introduceti orasul.
</div>
</div>
<div class="col-md-2 mb-3">
<label for="validationServer05">Strada si numar</label>
<input type="text" name="street_supplier" class="form-control is-valid" id="validationServer05" placeholder="Strada" required>
<div class="invalid-feedback">
Va rugam sa introduceti strada.
</div>
</div>
<div class="col-md-2 mb-3">
<label for="validationServer04">Banca</label>
<input type="text" name="bank_suplier"class="form-control is-valid" id="validationServer04" placeholder="Denumire Banca" required>
<div class="invalid-feedback">
Va rugam sa introduceti banca.
</div>
</div>
<div class="col-md-2 mb-3">
<label for="validationServer05">Cont</label>
<input type="text" name="no_cont_Bank" class="form-control is-valid" id="validationServer05" placeholder="Numar Cont" required>
<div class="invalid-feedback">
Va rugam sa introduceti numarul contului.
</div>
</div>
<div class="submit">
<button class="btn btn-primary" type="submit">Submit form</button>
</div>
</div>
</form>
</body>
</html>
2 個解決方案
解決方案1
0 2018-06-02 12:17:06
該查詢看起來很奇怪：
$qry=mysqli_query("INSERT INTO supplier `VALUES('sss','%$id_drugstore%',$_POST['name_suplier']','$_POST['county_supplier']','$_POST['county_supplier']','$_POST['town_supplier']','$_POST['street_supplier']','$_POST['bank_suplier']','$_POST['$no_cont_Bank']";`
- 刪除反引號`
- 缺少一個左
'前$_POST['name_suplier']' - 該字符串中的所有
$_POST變量都將永遠不會被替換，將它們放在花括號中：{$_POST['name_suplier']} - 確保那里沒有錯別字
name_suplier與town_supplier - 在查詢末尾缺少右括號，並且作為
mysql_query()括號
旁注：在該INSERT中也應該沒有字段列表嗎？
解決方案2
-1 2018-06-02 11:46:20
您的代碼應該是。 如果您的供應商表具有要插入的所有此字段，則可以正常工作，但是如果您跳過任何字段，則mysqli_query不起作用
$qry="INSERT INTO supplier `VALUES('sss','%$id_drugstore%',$_POST['name_suplier']','$_POST['county_supplier']','$_POST['county_supplier']','$_POST['town_supplier']','$_POST['street_supplier']','$_POST['bank_suplier']','$_POST['$no_cont_Bank']')";
$inserted=mysqli_query($qry,$conn);
使用 php mysql 在另一個表中插入選定下拉列表值的主鍵作為外鍵
[英]Insert Primary key of Selected Drop down list value as a foreign key in another table using php mysql
如何從下拉列表中捕獲主鍵並插入為另一個表的外鍵？
[英]How to capture primary key from drop down and insert as foreign key of another table?
根據另一個下拉列表的輸入顯示下拉列表時出錯
[英]Error while displaying a drop-down list based on input from another drop-down list
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