{"id":27619,"date":"2022-12-25T12:10:08","date_gmt":"2022-12-25T06:40:08","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-how-to-use-a-php-variable-in-a-query-to-select-something-where-thing-thing-closed\/"},"modified":"2022-12-25T12:10:08","modified_gmt":"2022-12-25T06:40:08","slug":"solved-how-to-use-a-php-variable-in-a-query-to-select-something-where-thing-thing-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-how-to-use-a-php-variable-in-a-query-to-select-something-where-thing-thing-closed\/","title":{"rendered":"[Solved] How to use a php variable in a query to SELECT something WHERE thing = $thing? [closed]"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-63472652\" class=\"answer js-answer accepted-answer js-accepted-answer\" data-answerid=\"63472652\" data-parentid=\"63472178\" data-score=\"-1\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<ol>\n<li>first thing i did escaping the string that&#8217;s going to be passed to the query to prevent SQL Injection but this might help but not giving you full protection consider using prepared statement&#8217;s.<\/li>\n<li>you are trying to output the <code>mysqli_query()<\/code> result as a string you cannot do that you have to fetch the data from the result to some kind of array then fetch it using <code>while<\/code> loop. to fetch those data you need <code>mysqli_fetch_assoc()<\/code> function.<\/li>\n<li>feel free to change the column name to whatever you want from clubID to clubName for example if you have column name named clubName you can also put them side by side if you want to output all column data but in your <code>select<\/code> statement you have to fetch <code>*<\/code> data to be able to do that any way copy and paste the following code and let us know what happens. {}this is to be able to use single qoute in the double qoute to fetch the data from the array if you don&#8217;t do that the variable won&#8217;t expand.<\/li>\n<\/ol>\n<p>.<\/p>\n<pre><code>&lt;?php\n    session_start();\n    $email = mysqli_real_escape_string($link,$_SESSION[\"email\"]);\n    echo $email;\n    $sqlget = \"SELECT clubID FROM users WHERE email=\"$email\"\";\n    $sqldata = mysqli_query($link, $sqlget) or die('error');\n    while($sqloutput=mysqli_fetch_assoc($sqldata)){\n        echo \"{$sqloutput['clubID']}&lt;\/br&gt;\";\n    }\n?&gt;\n<\/code><\/pre>\n<\/p><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\"><\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved How to use a php variable in a query to SELECT something WHERE thing = $thing? [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] first thing i did escaping the string that&#8217;s going to be passed to the query to prevent SQL Injection but this might help but not giving you full protection consider using prepared statement&#8217;s. you are trying to output the mysqli_query() result as a string you cannot do that you have to fetch the data &#8230; <a title=\"[Solved] How to use a php variable in a query to SELECT something WHERE thing = $thing? [closed]\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-how-to-use-a-php-variable-in-a-query-to-select-something-where-thing-thing-closed\/\" aria-label=\"More on [Solved] How to use a php variable in a query to SELECT something WHERE thing = $thing? [closed]\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[340,339],"class_list":["post-27619","post","type-post","status-publish","format-standard","hentry","category-solved","tag-mysql","tag-php"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>[Solved] How to use a php variable in a query to SELECT something WHERE thing = $thing? 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