php 显示 SQL 查询结果
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Display SQL query results
提问by jc70
I am having trouble displaying results from a SQL query. I am trying to display all images and prices from a products table.
我无法显示 SQL 查询的结果。我正在尝试显示产品表中的所有图像和价格。
I am able to display the echo statement "Query works" in the browser. But, the results are not displaying in the browser.
我能够在浏览器中显示 echo 语句“查询有效”。但是，结果没有显示在浏览器中。
if ($count > 0) {
echo "Query works";
} else {
echo "Query doesn't work" ."<br/>";
}
PHP Code:
PHP代码：
$con = getConnection();
$sqlQuery = "SELECT * from Products";
// Execute Query -----------------------------
$result = mysqli_query($con, $sqlQuery);
if(!$result) {
echo "Cannot do query" . "<br/>";
exit;
}
$row = mysqli_fetch_row($result);
$count = $row[0];
if ($count > 0) {
echo "Query works";
} else {
echo "Query doesn't work" ."<br/>";
}
// Display Results -----------------------------
$num_results = $result->numRows();
for ($i=0; $i<$num_results; $i++) {
$row = $result->fetchRow(MDB2_FETCH_ASSOC);
echo '<img src="'.$row['Image'].'>';
echo "<br/>" . "Price: " . stripslashes($row['Price']);
}
}
Screenshot 1
Screenshot 2: removed the images from the database, and used a filepath instead
截图 1截图 2：从数据库中删除图像，并使用文件路径代替
Screenshot 3: print_r($row)
截图 3：print_r($row)
回答by codaddict
I think
我认为
$row = mysqli_fetch_row($result);
$count = $row[0];
should be
应该
$count = $result->numRows();
if ($count > 0) {
echo "Query produced $count rows";
} else {
echo "Query produced no rows" ."<br/>";
return;
}
And your for loop should use fetch_assocas:
你的 for 循环应该fetch_assoc用作：
while ($row = $result->fetch_assoc()) {
echo '<img src="'.$row['Image'].'>';
echo "<br/>" . "Price: " . stripslashes($row['Price']);
}
回答by Nik
try
尝试
$sqlQuery = "SELECT * from Products";
// Execute Query -----------------------------
$result = mysqli_query($con, $sqlQuery);
if(!$result) {
echo "Cannot do query" . "<br/>";
exit;
}
$row = mysqli_fetch_row($result);
$count = $row[0];
if ($count > 0) {
echo "Query works";
} else {
echo "Query doesn't work" ."<br/>";
}
// Display Results -----------------------------
$num_results =mysqli_num_rows($result);
for ($i=0; $i<$num_results; $i++) {
$row = mysqli_fetch_assoc ($result);
//print_r($row);
echo '<img src="'.$row['Image'].'>';
echo "<br/>" . "Price: " . stripslashes($row['Price']);
}
回答by Alex
It's displaying characters because that is how you have stored the image. In order to show the image you are going to have to draw the image with something like:
它显示字符，因为这是您存储图像的方式。为了显示图像，您必须使用以下内容绘制图像：
echo '<img src="data:image/gif;base64,'.base64_encode($row['Image']).'" />';
echo '<img src="data:image/gif;base64,'.base64_encode($row['Image']).'" />';
回答by Jochem
$row is the first result-row (if any) from your query. $row[0] is the first column in this query (which, since you use select *, depends on the order of the columns in your database). So, whether $row[0] > 0 depends on the content of your database.
$row 是查询中的第一个结果行（如果有）。$row[0] 是此查询中的第一列（因为您使用 select *，这取决于数据库中列的顺序）。因此，是否 $row[0] > 0 取决于您的数据库内容。
回答by Fanis Hatzidakis
Mysqli doesn't have fetchRow(), that's part of the Pear::MDB2 library
Mysqli 没有fetchRow()，这是 Pear::MDB2 库的一部分
See the docs: http://www.php.net/manual/en/mysqli-result.fetch-assoc.php
请参阅文档：http: //www.php.net/manual/en/mysqli-result.fetch-assoc.php
Change your loop to the following:
将您的循环更改为以下内容：
while ($row = $result->fetch_assoc()) {
echo '<img src="'.$row['Image'].'>';
echo "<br/>" . "Price: " . stripslashes($row['Price']);
}
Also, by doing this:
此外，通过这样做：
$row = mysqli_fetch_row($result);
$count = $row[0];
before the loop you are essentially skipping the first row and not displaying its image in the loop.
在循环之前，您实际上是跳过第一行，而不是在循环中显示其图像。
回答by In81Vad0
to print all results from a query you can use a while loop
要打印查询的所有结果，您可以使用 while 循环
while($row=mysqli_fetch_assoc($result)){
echo 'Price '.$row['Price'].'<br/>';
}
回答by user1932459
Instead of
代替
If ($count > 1)
Try
尝试
If ($count >= 1)
